foreign [Music] okay welcome to this video on everything you need to get a grade six to nine in maths there is an absolute ton of content in this do make sure you check out my video that i did previously that i've linked in the description and everything to get a grade five you do need to still have a solid understanding of all those topics but here is a good accumulation of a lot of topics for eurovision if you are aiming for those higher grades in your math gcse obviously again you do need to conduct your own subject analysis just to make sure that you do cover everything off but there's tons here so we're going to get started just make sure you like the video share it with your friends and obviously please subscribe to the channel okay and obviously don't forget to follow me on instagram and on twitter as well okay so kicking this off a fractional negative indices so you've just got to remember looking at this one on the left if there is an indice the negative does the reciprocal which i always just write as a flip the bottom is the number underneath which does the root and then the number on the top is just a normal power okay so i'm just going to put power with that so normal power so for this particular one we've got all three going on so i always start with flipping it over doing the reciprocal and remembering that eight is just a number so we could write that as eight over one so when we flip that over it becomes one over eight then we've got to do either the root or the power i always like to do the root first make the number smaller before trying to then do the power so a 3 on the bottom would represent a cube root so we need to do the cube root of both of these numbers the cube root of 1 is 1 and the cube root of eight is two two times two times two is eight so it's one half then we've got to finish it off we've still got to do that normal power there if it was a one it wouldn't affect it but a two as a normal power is squaring so if we square them both we get one over four as a final answer and remember to follow all those rules now look at the one on the left we've not got a negative so we don't have to flip it over but we have got a 2 on the bottom which is the root again and a 2 is a square root and then we've got a 3 on the top which is going to cube the numbers so to start with the roots to start with if i do the square root of both of these and it's always a little hint there they are square numbers the square root of 9 is 3. the square root of 100 is 10 and then we've got to actually cube them as well so if i cube these as well let's write that in 3 cubed is 27 and 10 cubed is a thousand so it'd be 27 over a thousand just being very careful to get those right so that's negative and fractional indices okay so when you see this sort of question you see this these words here about bounds there's always something we need to write down straight away so we've been given two numbers here so s is correct to two decimal places and t is correct to one decimal place so the first thing i'm going to write down is the error intervals for them before i move any further so writing out error interval arrows pointing to the left and an equal to symbol on our lower bound there so i always start with the upper bound and i just stick a five on the end so eight point two four with a five on the end and then we just need to take that five off for the lower bound so it'll be eight point two not four but three with a five on the end and then do the same for the other one as well so let's write down the upper and lower bound for three point five so three point five sticker five on the end and then for the lower one three point drop it down to a four and again just stick a five on the end now this question here asks us to find the upper bound so when it comes to a fraction and you could have anything here but when it comes to a fraction if we want to get the biggest possible answer we want to do either we've got to decide here either we're going to do the upper bound divided by the lower bound or we're going to do the lower bound divided by the upper bound we're just going to think which one there is going to give us our biggest possible answer and that is the upper bound divided by the lower bound so we want the biggest number on top divided by the smallest number and that'll give us our biggest possible answer so we want the upper bound of s s sitting on top so the 8.245 and the lower bound of t 3.45 sitting on the bottom so this is a calculator question so we just need to type that in the calculator 8.245 divided by 3.45 and see what we get as our answer there so if we type that in 8.245 divided by 3.45 and we get the answer it's quite a lot of decimals here two point three eight nine eight five five zero seven always just write down all the numbers and then round it however you've been asked in the question this one says three decimal places so let's chop it after the nine and that would go to two point three and this is a bit of a dodgy one because the eight rounds the nine up so go to two point three nine zero keeping the zero in there it does say three decimal places so two point 2.390 would be my answer there okay obviously it might say lower bound as well in which case you'd use the other side of the formula would you use this one instead you might even be asked to do both of them and then round it depending on how what they both look like so if we imagine that the lower bound comes out as if we just make something up here let's imagine the lower bound comes out as 2.295 or something like that now it might say considering bounds give your answer to an appropriate degree of accuracy and if you look at both these numbers if we just imagine this is the lower bound it's probably not but if we imagine it was these two numbers would not round the same to two decimal places not to three decimal places not even to one decimal place because the one on the left there would round to two point four and the one on the right around to two point three so if i just have to consider bounds to give an answer for this i'd just have to say two and i would say the answer is two and that would be to one significant figure so i could be asked that as well and i just look at the upper and the lower bound and around according to um how they both actually round and this these in this case my little example here they only round the same to one significant figure okay so recurring decimals as fractions so a little rule we are going to use a bit of algebra for this um but if there's two recurring decimals we're going to times it by a hundred and if there's one recurring decimal times it by 10 and so on there's lots of different ways of doing this this is just the way that i do them so i write what the decimal is so we've got x equals 0.54 and remember if there's a dot above the 5 and the 4 that means that that decimal there is 0.5454 and continues like that it's the 5-fold that's recurring it's always good just to write that down just to remind yourself what that decimal actually represents so if we've got two recurring decimals we can times that by a hundred you'll see why if you've not seen this before hopefully it's just a reminder you get 54.54 now we can take them away from each other and that'll eliminate these recurring decimals here if we take them away from each other we have 99 x's equals 54 and being make make sure that you actually do use your algebra to show this now we can divide both sides by 99 so x equals 54 over 99 and there it is as a fraction you do just need to just check just to see if your fraction simplifies so do check to see if this one simplify this one actually does the top and the bottom both divide by nine and you'd normally be asked to show that it equals this next one but if it doesn't you know you can look to simplify if it says to give it in its simplest form and if you divide them both by 9 you get 6 over 11. so the question might have said show that it equals 6 over 11 and that's how you would show it you'd show how it simplifies now onto the one below it's only got one recurring decimal but it's a bit of a nasty one um so if we think about what this actually this decimal actually looks like 0.23 with the dots above the three that's naught point two and then a recurring set of threes it's not a full repeating pattern of two three two three two three so if we have a look at this one and again there are different ways of doing this type of question here but this is i just followed the same process for all of them so x equals naught point two three with the dots of the three and there's one recurring decimal so let's times it by 10. so we get 10x equals 2.33 and then the 3 keeps going i'll just balance it out so it's the same amount of decimal places in both now if we take these away from each other something kind of interesting happens here we get 9x equals and 2.33 take away 0.23 is 2.1 now if we take the same approach as before and turn that into a fraction so dividing both sides by 9 we get x equals 2.1 over nine and obviously you can't have decimals in fractions so we need to remove this decimal from the top so we can times the top and bottom by 10 and that would give us 21 over 90. there you go and there's your fraction and again you just need to check to make sure that's if you see it to see if that simplifies or not and actually does the top and bottom both divide by three so you can divide the top and bottom by three and you would get seven over again so it might have said show that this fraction equals 7 over 30. but always just check to sit at the end see if it simplifies obviously if it just said to write it as a fraction 21 over 90 would be fine but if it said to give it its simplest form we would have to simplify there okay so some compound interest james invests two and a half thousand pounds for three years in a savings account he gets three percent per annum compound interest in the first year and then x percent for two years he has this much in the at the end of the three years work out the value of x well just like normal compound interest here we can actually do the first year so we can work out i'm just going to label this up so after the first year let's see how much money he has so 2 500 for 3 we would times that by 1.03 obviously we don't need to add any powers into there because it is just the first year that he's getting this so if we do this on the calculator times 1.03 and we get an answer of 2575 pounds now we've got a bit of a reverse compound interest scenario we have to create a little formula for this because in the second and third year he gets x percent now his starting number and let's just write down second and third his starting number now is 2575. now if we knew the percentage we'd multiply it by that percentage which we'll call x in this scenario because we don't know what it is and we do that to the power of two as it's two years and what it's saying in the question is that gives us the answer 2705 pounds and 36p obviously at this point here we could just guess some numbers we could just guess an x value and if it's too high go lower if it's too low go higher and that's a fine way of doing it as well but you can save yourself a bit of time here just guessing lots of numbers by actually just rearranging this so we can divide both sides by 2575 and it will give us the value of x squared so x squared is going to equal and again let's just use on the calculator 2705 36 divided by that gives us this answer here 1.0506 25243 that's obviously the value of x squared so we need to square root both sides so if we square root that answer we get the answer one point zero two five zero zero zero one one eight so that gives us a little percentage here so it's hidden within this decimal now our actual percentage here is not 1.025 because that one point is what we increased by for the percentage so it's just this zero two five that we're interested in and zero two five if you think back to what we did at the start here with zero three being three percent well zero two five would be two point five percent there we go so that's two point five percent so just be careful you do read that off there and don't put one point zero two five in this type of question but again you could have guessed if you'd guessed one point zero 1.02 it's normally going to be lower than that first one that's normally what banks do they give you a better one in the first year and they're not so good following that okay so there's just what two ways that you can approach that you can guess some numbers or you could uh obviously just rearrange it and find it there just be very careful that you actually read the percentage from that one point number okay so we've got some sirds questions so 3 plus root 5 squared in the form a plus b root 5 where a and b are integers okay so we need to have a look at this so that obviously is a double bracket there so if i rewrite this as a double bracket we've got three plus root five and three plus root five and we just need to expand that now and see what we get so lots of little third rules going on in this one so three times three gives us nine three times root five is three root five five root five times three is another three root five and then root five times root five is root twenty five and it just becomes five remembering similar simulators when all these exact same serves when they when you multiply them together become the whole number and if we simplify all this down 9 and 5 makes 14 and then we can add together these thirds in the middle because they both have a root five so we've got three lots of root five and three lots of root five which is six lots of root five and there's your answer just there so a would be fourteen and b would be six and that is in the form a plus b root five and there's our answer with a bit of thirds so look at another third's question okay so we've got show that this can be written in this form now when we've got thirds on the bottom of a fraction we're always going to look to rationalize it now with some of the easier thirds you just times the top and bottom by whatever this root is just here so by root 3. unfortunately that's not going to work for this one so we've got 2 plus root 3 and if we did times the bottom by root 3 there we'd get 2 root 3 plus 3 and we'd still have a third on the bottom so what you do when you've got this on the bottom is you times it by the denominator still but we change this little sign here so whatever that sign is we're going to flip it to the opposite so we're going to times it by 2 minus root 3. so always write that down okay because you do get marks for these these uh elements that you're working out so i'm just going to rewrite the fraction 2 plus root 3. i'm going to show next to it what i'm timing the top and bottom by so i'm going to times by 2 minus root 3 and i'll put that in a bracket because i'm going to do a double bracket here and i'm also going to times the top by 2 minus root 3 because we're creating an equivalent fraction here so it's up to you if you want to put these in brackets as well just to remind yourself you're doing a double bracket and then you just need to multiply both of them out and see what we get so on the top and i'm not going to draw all over it i'm going to times them both by the 5 first and then times them both by the 2 root 3. so 5 times 2 on the top gives us 10. 5 times negative root 3 gives us negative 5 root 3 on to the 2 root 3. so 2 root 3 times 2 is positive 4 root 3 and then 2 root 3 times negative root 3 and this is the multiplication you always just got to be careful of gives you 2. i'm going to do it to the side we get 2 lots of root 9 2 lots of root 9 is 2 times 3 which is 6. so that number at the end there and careful because 1 positive 1 is negative so we get negative 6. i'll tidy that all up in a sec let's have a look at what's on the bottom now the whole reason we're doing this on the bottom is to eliminate the thirds in the middle so you can do a bit of a trick when you're doing these you can just do the first two which is two times two which is four and the last two so root three times root three is three and it's minus 3. there we go so that's why our fraction is disappearing because it's 4 minus 3 on the bottom let's tidy up what's on the top so on the top there we have 6 and the 10 so we've got 10 minus 6 which is 4. we've got negative let's do this in a different color we've got negative five root three add four root three negative five add four is negative one so minus one root three or minus root three and that's all over four take away three so it's over one so when we're dividing by one we can get rid of the fraction there and we get four minus root three so there's some third elements to be thinking about obviously there are so many different questions on thirds that you could actually have there's a couple of things there to be thinking about the other thing is obviously to remember that sirds can only be collected together like in here when the number underneath the third is is the same okay so you will have to know how to simplify thirds as well but i've got plenty of videos on that to check out okay so we've got some standard forms we've got some multiplication and some division i'm going to start with this multiplication now i'm primarily thinking this is like a non-calculator style question because obviously if this was on a calculator you could just type it in nice and easy and write it in standard form so actually we're going to have to do a bit of a trick here so 3.2 i can times that by 4 to figure out what my starting numbers would be so 3.2 times 4 again you've just got to work that out whatever method you're used to using i'm going to do 32 times 4 and hop the decimal back in so 4 times 2 is 8 and 3 times 4 is 12 and then hop the decimal back in so it's 12.8 so the answer i get there is 12.8 it's going to be times by 10 and when we're multiplying we can add together the powers here so 3 plus 4 gives us 7. obviously though it says to write your answer in standard form and that number here 12.8 is not between 1 and 10. so if i hop the decimal in making the number one place value smaller i'd get 1.28 and if i make the number smaller i have to make the power one bigger so it'd be 10 to the power of 8. okay so don't be put off for these sorts of questions do your multiplication follow your power rules and then balance it out remember if you make the numbers smaller you make the power bigger i'll take this exactly the same approach for this one except it's a divide so i need to do 3.6 divided by 4 and again it's up to you you can hop the decimal out and hop it back in but i'm just going to do 3.6 divided by 4 using a bit of bus stop so 4 doesn't go into 3. so carry the 3 and then 4 goes into 36 9 times there we go so it's 0.9 so when we divide that we get 0.9 times 10 to the power of and when we're dividing we can subtract the powers so the first power there is 2 the second power is negative 3. so just write that down to the side what we've got we've got 2 take away negative 3 which is to add three which gives us a power of five there we go so the power is five and then again we just need to balance that out because obviously this is not between uh one and ten so i'm gonna hop the decimal this way making it nine so this time i've made the number bigger so i need to make the power smaller so it goes down to a four and there's our final answer nine times ten to the four and the last question on this section so we've got in 2008 alice bought a house in 2000 uh sorry in 2008 alice bought a house in 2014 she sold it for 20 profit and in 2019 it was sold at five percent loss and it gives us the value that she sold it for work at the cost of the house in 2008 so right back at the start so we had 2008 she bought a house and she sold it for a 20 profit then in 2014 um it was sold for a five percent loss and then in 2019 the value of it was 182 thousand four hundred right so at this point in 2019 this cost here to whoever she sold the house to in 2014 this was sold at five percent loss so to her that was only 95 percent of the value that she bought it for so if you've got a calculator here um in order to turn that back into a hundred percent quite nice and simply we can divide it by 95 to get one percent so if that there is 95 we can divide that by 95 so divide by 95 which would give us one percent so one percent would equal and we'll just do this on the calculator 182 400 divided by 95 gives us 1920 and then to get back to a hundred percent we just times that by a hundred so we can have just add two zeros on for that so one nine two zero with an extra two zeros there we go so a hundred and ninety two thousand so that's the price that it was sold for uh sorry that she paid for it before so before that it was 192 000. that's not a zero 192 000 and then we've got to think about the original sale in the first place so that means um that that 192 000 there was a 20 profit so if that was a profit to alice in the first place then to her this amount of money was 120 okay it was 20 more than she originally paid for it so we can follow exactly the same approach now if that's 120 we would divide that by 120. to give us one percent so one percent here would equal and we'll just work that out divide that by 120 and one percent is 1600 and again you can times that by a hundred to get a hundred percent and our original cost up here would be one six zero zero with two zeros on the end so 160 000 pounds there we go so 160 thousand there we go so that's it using a calculator that's moving through two reverse percentages there one with a decrease one with an increase just remember if something's been decreased then we get this 95 percent whatever was less left over and then there was also the scenario where at one point it was increased making it 120 percent that extra 20 there so this can be done calculator and it can be the non calculator the 95 one wouldn't be very nice but thinking about how you could have done this 120 which is quite common when v8 is added on because they often have these v8e questions or they say v80 was added on at 20 there is a way of doing that without a calculator and what you could do when that happens is you can divide by 12 and then times by 10. so you could do that without a calculator you could do a bit of bus stop there and divide 192 000 by 12 and then just add on a zero many times by 10. okay there we go okay and on to algebra this is a bit of a long one let's have a look so we've got triple brackets to start with now you can expand any two that you want to start with i'm just going to go for the first two so if we expand those to start with let's have a look at what we get so two x plus three multiply by x plus four now if we expand that i'll just following normal expansion rules we get 2x squared plus 8x plus 3x plus 12 on that last bit there okay tidying that up first so tidy this up simplify it down we get 2x squared plus 11x plus 12. now we can multiply it by the final bracket the final bracket being up there the 3x minus 1. so stick this in a bracket and multiply it by three x minus one again just take your time with this it does get quite messy when you're doing it okay so just spend the time make sure that you get each piece perfect so let's move this out of the way and let's just be really really careful when we expand this now so 2x squared times 3x gives you 6x cubed and then 2x squared times minus 1 is minus 2x squared onto the 11x let's go for a different color so 11x times 3x is 33x so positive 33x squared again there we go and then 11x times negative 1 is a negative 11x two more to finish this off 12 times 3x is 36x just about fit this in and then 12 times minus 1 is minus 12. right there we go we just need to tidy all this up now because we've got a lot of pieces there we've got the 6x cubed at the start we've got negative 2x squared add 33 x squared so negative 2 add 33 is 31 so plus 31x squared negative 11 add 36 is 25 so 25x squared sorry 25x and then the minus 12 at the end and then do just go back double check make sure you've added up the pieces correctly uh just to make sure you don't make any mistakes on that final part there that's expanding a triple bracket just remember to tidy up your first double bracket and then multiply it very carefully by the next one okay so factorizing a quadratic and here we've got a coefficient bigger than one at the start so you need to think about how you do this you can have two three five you can have any combination of numbers here but in order to get a two at the start there we're going to need two x in one of our brackets and the next in the other and i always just follow the same process write down the factors of 12. so 1 and 12 2 and 6 3 and 4. and we've just got to think okay well one of these numbers is going to double and we want to make negative 5. obviously this one has a negative 12 at the end so one's going to be positive one's gonna be negative and when one doubles we're trying to make that number five in the middle so it's not gonna be one and twelve because we can't make five particularly if we double the twelve we're not going to make five if we've got a 24 and one we're not going to make five out of two and twelve so it's not that one let's have a look at the next one doubling the four two would make four so four and six can't make five out of four and six doubling the six would make twelve and you can't make five out of um out of two and twelve so that's not gonna work either let's try the three and four double the three it'd be six we can't make five out of six and four so it's gonna have to be this last one doubling the four which is going to make eight so three and eight and that works she could do positive three take away eight that'll give us negative 5. so we want the 4 to double so we stick that in the opposite bracket to the 2x we want that one to be negative so stick a negative with that and stick your plus 3 in the other one and there is your factorized double brackets just remember as well it could say actually to solve something like this and you can solve by factoring if it did say the word solve then we'd have two solutions here for this right bracket it's nice and easy just flip the sign we'd have x equals four and for this left bracket here nice and easy again it's still flipped the sign still three but because of this 2x here it'd be minus 3 over 2 or minus 1.5 so if it did say solve that be an additional step that you could go about you could say x is negative 1.5 and x is 4. okay okay so we've got rearranging or making extra subjects here and we've got two x's in our formula so i mean we can have so many combinations of this question that you can have the whole idea of this type of question though is that you get the x's on the same side of the equal sign everything else to the other side obviously the x's are on the same side of the equal sign at the moment we want to get rid of this fraction so you might not have a square root in a question but in order to get rid of this square root to start with we need to square both sides so if we square both sides we get y squared equals and then what's left under the square root it's not going to change that we're just going to have x plus 3 over x minus p now again we've still got a fraction involved so we need to get rid of this denominator so whatever's on the bottom there we're just going to times both sides by what's on the bottom i'm going to stick it in a bracket because it's easy for me to move it so x minus p and times the other side by x minus p okay so we've got a bracket on the left we've got y squared brackets x minus p equals x plus 3. now as soon as we got rid of this bracket we can start to actually move the pieces out of it so if we expand the bracket we get y squared x minus y squared p or py squared however you want to write it and that equals x plus 3. now i can start to move the pieces either side so let's go for there's two ways of doing this one at the moment i could move the x's to the left or the x's to the right now what i'm going to do is i'm going to move the x's to the left so i can get rid of this negative y squared p or plus p y squared i'm going to rewrite it differently so plus p y squared to both sides let's bring this up here and we get y squared x or x y squared i just probably should have rewritten it there it doesn't really matter add the over so we get x plus 3 plus this new p y squared over here now we can minus the x from both sides we can get all our x's on the left so minus x so and again i will write this style i'm going to write x y squared i just prefer it x y squared minus x equals three plus p y squared and here's where this process starts to come in now we can factorize this left hand side so that we get one x so we'll get x brackets y squared minus one that equals three plus p y squared and then you can divide by this bracket here and slot it underneath the right hand side so you get x equals three plus p y squared over y squared minus one and that would be our final answer obviously there is an unlimited amount of different formulas that you could have to rearrange so i'm only going to go over the one here because it involved a lot of different steps there getting rid of a square root by squaring getting rid of a denominator by timesing it across putting it in a bracket in this second step that was over here and then we had obviously expanding a bracket getting all the x's on the same side factorizing it to finish i mean you've got literally every possible step you could think of in this one and so it's quite a very quite difficult one there but there's just an example of some rearranging and what to do when there's two uh subjects that you're trying to turn into the one okay so working out the nth term for this sequence so whenever you're working at the nth term of a sequence this is a quadratic one it doesn't say in the question but it is a quadratic you hopefully can see that just by looking at it because here it goes up by four and here it goes up by six and then here it goes up by eight okay so because of that we've got a second difference here so the second difference is two now whenever you get a second difference what you need to do is actually divide that second difference by two it's just something which happens with the square numbers if you think about the square numbers and this is why it's a quadratic sequence 1 4 9 16. the second difference comes out as 2. let's have a look so 3 5 7 and the second difference is 2 and that tells you it's related to the square numbers but it can be more than just the square like one lot of the square numbers so if we divide this by two it tells us it's one n squared or it's related to the end the square number sequence which is this one here n squared equals 1 4 9 16. so what i always do is write down the square numbers now again i know there are formulas for this but i just really prefer this method so 1 4 9 16. but now if you have a look there is actually a difference between the n squared sequence and the one going up there in the question to get from 1 to 15. let's just write this down what's the difference to get from 1 to 15 that's plus 14. to get from 4 to 19 that's plus 15 and to get from 9 up to 25 that's plus 16 and this just carries on look from 16 to 33 is plus 17. so not very nice look if we had a constant difference if they are all plus 14 then you'd just put n squared plus 14. but unfortunately they're not so we can't do that they're not all just plus 14. so when you get this scenario here you've got to have a look at this sequence and think okay well i'm going to have to just work out the nth term of that sequence and the nth term of that sequence is it goes up in ones look 14 15 16 17 it goes up in one so one n or n and then it's obviously not the one times table or if you you know the little trick where you go back here you take away one you'd get a 13 at the start so it'd be plus 13 because it's 13 bigger than the one times table so that sequence there has an nth term of n plus 13. so all you do is just slot that at the end of n squared so n squared plus n plus 13 and that would be your nth term of this quadratic sequence here okay n squared plus n plus 13. you can get something slightly different here like if you had a second difference of if we just imagine this to the side if you had a second difference of four so if it was four your sequence would start with two n squared and then you would have to write out double the square numbers so instead of writing 1 4 9 16 you would write 2. what happened there you would write 2 8 18 32 and then you would compare the difference back to these numbers similarly if it was 3n squared if there was a second difference of 6 you would triple all the square numbers and compare it again obviously just remember if this comes out as a constant difference down here that's great you can just put that after the n squared part or the 2n squared but if it is a new sequence you are going to have to forget about the rest of the question for a second and just work out the nth term for that sequence and then slot that on the end so i'll finally answer that n squared plus n plus 13. okay so this question says find the coordinates for the turning point of the graph with this equation so to find the turning point we have to do something called completing the square and sometimes it might say write it in the form it might say x plus a squared plus b it might say write it in that form okay and that just means completing the square and it's how we find the turning point so in order to do this you have a look at the x coefficient which is there and we have that number and we stick it in the start of our bracket like this so we have x plus eight half this four squared and the reason you do that is because if we expand this bracket if we think about it to the side we expand that bracket there x plus four x plus four x plus four there we go we would get x squared plus four x plus four x which will give us eight x but we would get plus sixteen at the end so obviously it's similar to this part just here we've still got the x squared we've still got the eight x but we've got plus 16 we don't want plus six then we want plus three because what we're doing is we're rewriting it as in in a different form in this completed square form so in order to make sure 16 becomes three we would have to subtract 13 from this double bracket there 16 take away 13 give us three so that is actually written in completed square form there and that is how you do it now you don't have to fully expand the double bracket you can just think okay well four times four sixteen that's the number i know i'm going to get at the end so i'm gonna have to take away 13. and our coordinates for the turning point are written within this so the x coordinates in the bracket there just like when you're solving you flip the sign so the x coordinate is negative four instead of positive four and the y coordinate here is just this number at the end no flipping of the sign just writing whatever's there so minus 13. there goes the coordinate of our turning point is negative 4 and negative 13 for the y-coordinate and that's completing the square okay a longer question here we've got solve the simultaneous equations now we've got an x squared and a y squared this actually represents the equation of a circle and the equation of a line okay and that line would be good slope in this way i believe let's have a look okay so when you get this sort of scenario here you get two solutions because there's two crossover points there is one scenario where you might get one solution which is where it's a tangent to the circle if it just touches at one point there could even be no solutions but it wouldn't be unlikely we'd be asked to try and solve something with no solutions so if we have a look at this we've just got to basically sub one into the other and we can sub this x minus y equals two into the equation of above so if we actually do that first of all we've got to rearrange it now there's two ways i could make y the subject or x the subject so i'm going to make y the subject so i'm going to add y to both sides uh which actually that would straight away make x the subject so let's do that y equals sorry x equals 2 plus y okay you could even write that as y plus 2 if you prefer and i'm actually going to do that x equals y plus 2. now what we can do is we can sub this little expression here y plus 2 in place of the x in the first equation this x squared so if i just do that so rather than writing x i'm going to write y plus 2. so y plus 2 squared plus y squared equals 34. right so we've now got an equation that has no x's in and we know that when we've got an equation that only has one unknown in it we can definitely solve it um so if we actually go about expanding this expanding this first bracket here i'm not going to write it as a double bracket i'm just going to go for it so y squared will get plus 2y plus 2y so plus 4y and plus 4 at the end and then obviously we've got plus y squared equals 34. now we can tidy that up look because we've got these two y squared pieces so we can add those together just to make it a bit more tidy so we've got 2y squared plus 4y plus 4 equals 34. and now we can only actually solve one of these it's a quadratic so we can only solve that once it equals zero at the end so we need to minus 34 from both sides so if we minus 34 we get 2y squared plus 4y and 4 minus 34 is minus 30. and that now equals zero so we can actually go about solving this now now if in this point here we can i mean if we've got a calculator we could potentially use the quadratic formula we could factorize it from here but if you notice that all these numbers here all divide by something we can simplify it down because we can divide both sides by anything because the right hand side is just 0. it doesn't change it so if we divide everything by 2 we get y squared plus 2y minus 15 equals zero and it potentially just makes it a little bit easier to factorize there now if we go by actually factorizing this both are going to have a y in and we're just thinking what numbers there multiply to make 15 so we've got 3 and 5 or 1 and 15. i'm not going to write down the 1 and 15 because i've already spotted it's 3 and 5. and in order to make 2 we need plus 5 minus 3 and that's going to give us positive and positive 2 in the middle so we've got two solutions going on here obviously it's just these numbers in the brackets but flip the symbols so we've got y equals negative five and we've also got y i just want to write x there y equals three so there's our two y solutions we now just need to find our x solutions now we've got a nice little formula we made right at the start here to find x we've got x equals y plus two so when y is five x will equal negative five add two which is negative three so x equals negative three and then for the next one we've got y equals three so x equals three plus two so x equals five there you go so you've got your two solutions and they're written in pairs there we've got our first solution here x equals negative three y equals negative five and we've got our second solution over here x equals five and y equals three okay so there would be coordinates there so it might be if we think back to this graph it might be that this coordinate where it crosses over is negative three negative five and this coordinate up here is three five okay so that's what it represents there when we're solving these simultaneous equations and remember that could be solved graphically as well you might be given a diagram drawing on a straight line and then just you know estimating them or finding them exactly where they cross over okay so they're solving simultaneous equations and again there's loads of different variants here you might have um two that have y squared and a y in and you can set them equal to each other but you know this is just one of the harder ones here where you've got to sub one into the other okay so we've got some quadratic inequalities so we're going to solve this now obviously these um represent a parabola a curve a quadratic graph again a quadratic graph here normally when we're solving them we are solving to find these two roots here or solutions and you can be asked to represent uh to find them or to estimate them from a graph if you've been given a picture they're referred to as the roots but when we're solving an inequality we're looking at something slightly different now this first one here they're both exactly the same we're gonna have a look at the two different scenarios this first one says less than zero so essentially it's saying is what part of this curve is less than zero if you have a look this part of the curve is what's below ground what i like to refer to a less than zero z and b zero being the x axis there so really all we're going to say is well what part of the curve is between these two points below zero so if we actually just factorize this to find out what these two roots are so put it into a double bracket again it might have a coefficient bigger than zero but i'm just going for a nice easy one here just to show you the process so the factors are 24 we can have four and six two and twelve three and eight so i'm just gonna not write them all down because i've already spotted it's three and eight but that would be plus three minus eight and that would give us minus five so our roots here our solutions are x is negative three and x equals eight so if we were to label those on the graph here this would be minus three on the x-axis and this would be eight now the bit is below the curve is between negative three and eight so i'd have to write this as um one of these types of inequalities okay well we say it's between two points and that would be my answer there x is between negative three and eight or the part the graph that is under zero or less than zero is between those two numbers now that would change slightly for this one down the bottom because i've just realized they're pointing the same way that wasn't meant to be pointing that way let's change that that's meant to say bigger than zero okay so there we go sorry about that let's change that x squared minus five x minus 24 is bigger than zero so this would be something slightly different now because the bit that's bigger than zero and if i just redraw this graph there we go eight here negative three here the bit that's actually bigger than zero we've got one part over here and we've got a completely separate part over here so we've got two separate parts of the line and that means i can only write this as two separate inequalities now this part on the left is all this part of the curve is going to numbers less than three so my solutions there would be well anything that's less than three sorry negative three so x is less than negative three and that's one of my solutions for that part of the line but i've also got the other part and that is going to all the numbers bigger than eight it's going this way so my other solution would be x has also got to be bigger than eight so i've got two two ways of two solutions here to write with my inequalities i've got all the x values less than negative three and all the x values bigger than eight so that's solving a quadratic inequality just gotta think about which part the graph are you actually looking at okay and we've got a question here that shade the region that satisfies all of these inequalities and we've got these three inequalities there now some of these are quite easy to draw but when we're drawing these inequalities if we just imagine this to start with the the inequality is an equal sign it's just going to help us to actually draw the graph so what i'm going to do is i'm just going to imagine that this one says y equals 2x plus 1. i'm going to draw that line x plus y equals 4. i'm going to draw that line and i'm going to draw the line y equals one now the easiest one for us to draw there is y equals one now looking at my graph here you can't see the number one but it's between naught and two now on the y axis y equals one is at this point here and that line y equals one is just a nice straight line going through y equals one so that's our first one drawn there done the next one is x plus y equals four i'm going upwards now so x plus y equals four well if x plus y equals four that's quite a nice line to draw because that means when one coordinates zero the other has to be four so zero on the x axis gives us four on the y axis and zero on the y axis gives us four on the x axis so you've got a nice diagonal line there to draw between four on both the axes and these x plus y equals a number always just connects up the y axis and the x axis to those numbers so we've just got to do a nice neat diagonal going through with a ruler and a pencil and i'll continue that all the way oh that's not very good i'll continue that all the way to the top of the graph there just a nice diagonal going all the way through so that's that one done and the next one there is y equals two x plus one now you might know little tricks for how to plot these if not then i would suggest just drawing yourself a little table and picking some of the values you probably get away with doing two or three values but if we sub in um some x values to find our y coordinates here when x is zero that would be two lots of zero add one y is one so if i do this in a different color so when x is zero y is one that gives us that coordinate there let's pick two or three let's just go a little bit higher so when x is two two times two is four plus one is five so when x is two y is five and that is here and we could give some negative numbers as well now but hopefully you can get a ruler and a pencil and join those up now it has a gradient of two so i can just look at that coordinate on the um on the y axis there and go across one up two which gives us that coordinate there cross one up two gives us the one we've already drawn and i can follow that pattern backwards as well but you can draw a full table out for this just to make sure you get it's perfect but there we go there's my line i just need to connect those all together nice and carefully with a ruler and a pencil and there's our three lines drawn okay now it's a shade the region that satisfies all these inequalities now careful because it's not always that little triangle there that's made in the middle so i'm going to get rid of these equations now so i don't need them anymore but i need to have a look at the inequalities now all i'm going to need to do is pick a coordinate and test it in these inequalities so if we do pick this one in the triangle here so this coordinate here let's do it a different color there's a coordinate inside the little triangle in the middle which is the coordinate 1 2. this coordinate here well x is 1 and y is 2. and if i sub them into these inequalities as long as they all work then that's the region i'm going to shade so if i sub it into that first one there x is one and y is two so y e y is which is two is less than two lots of one which is two plus one so 2 is less than 3 and that works that's correct moving on to the next one then that one works we've got to test all three though x plus y is bigger than four so x is one so one plus y one plus two is bigger than four no that doesn't work so it can't be inside that triangle there although what we have figured out is that the first one works so in terms of that line y is less than 2 x plus 1 is my pink pinky red colored line there so it's got to be on the right hand side of that so it can't be this region on the left it can't be this region on the left and it can't be this region on the left so it's got to be one of the other four but it's not inside that triangle now so we can put a cross in there as well we know it's not that one so we've got three left to have a look at so if we get rid of this and obviously always leave you working out down but let's try another coordinate so let's try one in this region here as this one's on the top so that coordinate there is the coordinate for three okay so this one here four and three and if we test those out into all our inequalities and see if they actually work so looking at the first one uh well we've got y is three so three is less than two lots of x so two lots of four eight plus one so three is less than nine and that's correct that works moving on to this x plus y is bigger than four so x is three so three plus four is bigger than four well that's absolutely correct seven is definitely bigger than four so it works in that one as well and then on to the last one here y is bigger than three sorry y is bigger than one the value of y is three there and not on that coordinate so three is bigger than one perfect they all work so it's got to be that region there so we just have to shade this region in it might say to label it r or something like that but shade it in and there's your region done so once you've drawn your lines just pick a coordinate test out into your inequalities and as long as they all work and they're all correct then you can shade that region but do be careful you don't just shade the region of the shape there like the triangle in the middle because it's very often not just the shape that's formed okay so we've got solving a quadratic here and it says give your answer correct the three significant figures now this gives you a massive hint here three significant figures that it doesn't factorize and when a quadratic doesn't factorize when we've got to solve it with whatever scenario we've got to use the quadratic formula so one that you need to remember you've got minus b plus and minus the square root of b squared minus 4ac all over 2a and what i do straight away is i write down the values of a b and c so a is 5 the one in front of x squared so a equals 5. we've got b equals and you got to be careful because you've got to get the symbols in the signs in with it negative 11 so b is negative 11 and c is again that number at the end negative 13. so first things first we just need to put them all in the right place so minus b will be well negative b and b is already negative so that becomes positive so b is 11 plus and minus the square root of b squared and we're squaring a number here and it's negative so stick it in a bracket minus 11 squared take away 4 times 5 times negative 13 4 times 8 times c so 4 times 5 times negative 13. there we go you can stick that all in a bracket if you want just depends on what calculator you're using and that's all over 2a which is 2 times 5 which is 10 let's just write 2 times 5. now if you've got a casio calculator which hopefully you do because they are the best ones for these higher gcses then you need to just type that in as you see it so precious pressure fraction button 11 let's do plus i'm going to do the plus version first so i'm just going to do it once using the plus sign here and then i'll do it again in a different color using the negative so 11 plus the square root of in brackets minus 11 close brackets squared take away in brackets 4 times 5 times negative 13 go on to the bottom of the fraction and i'm just going to type in 10 i'm going to type that in turn it back into a decimal it says i get one which is 3.05192213 and then i'm just going to go back into my calculator change the plus sign for the minus sign and we'll write down what we get there so back into the calculator takes a bit of a while to get all the way back through this but it's better than typing it all back in again delete the plus put a minus press equals again and i get the answer negative naught point eight five one nine two two one two nine six there we go lots of digits there so obviously they're making a two solutions one with the positive there one with the negative then we've just got around it so it says three significant figures so in the first one there three zero and five are my first three significant figures so that would be 3.05 for my first solution and the second one's a little bit iffy here because we've got a zero at the start so it's one two three after the one so that'll be naught point eight five two because it's a nine after it there you go and there's my two solutions written to three significant figures just be very very careful when you're doing the quadratic form we know that you do make sure uh of a couple of things that is one when b is negative make sure it flips this sign for the first one and the second one is make sure you put these negative numbers in a bracket otherwise it's gonna it's gonna it's gonna give you an answer of negative 121 there rather than the positive 121. well if you are squaring a negative number you could just put 11 squared and forget about the negative but a little bit of a trick there okay there we go on to the next one whenever we get an algebraic fraction though we need to just need to factorize everything that we see no matter what the question whether your time is in dividing it's just simplifying just factorize everything and that can be quadratics with coefficients bigger than bigger than one we can have difference of two squares we can have single brackets but just look and see what can i do to factorize it so the top there there's no number at the end so it definitely doesn't go into a double bracket but the top of the 3x squared and the 9x both divide by 3 and they both have an x in so we can simplify it we can factorize it that way so 3x times x gives us the 3x squared and 3x times 3 gives us that 9x there so that's that factorized on the top and we have a look at the bottom that has a number at the end but no x pieces in the middle so that's the difference of two squares the factors are 9 it's the three and three one positive one negative that's what cancels it out so we have x plus three and x minus three there we go and when we've got these common brackets on the top and bottom we can cancel them out so the x plus three and the x plus three we can divide the top and bottom by x plus 3. so we end up with a final answer of 3x over x minus 3. and again you can have lots of different scenarios where you're simplifying an algebraic fraction again there could be different sorts of quadratics or just single bracket factorizing that you're doing but there is the idea of how you simplify these and no matter what question you see an algebraic fraction you're normally always looking to factorize look at one more okay so we've got some more algebraic fractions here now we're adding these together so we need a common denominator so i've got to solve an equation here but i've got to get these algebraic fractions added together first before i can do that so if we times both sides or this left-hand fraction top and bottom by x plus 1 and also times the top by x plus 1 that's going to allow me to get a common denominator as long as that times the right by the left denominator as well which is 3x minus 2 just following normal fraction rules there so times that by 3x minus 2. okay and i've written that next to the top and bottom so i don't forget what i'm times in by so on the top there we get eight lots of x plus one and we're going to add to that six lots of three x minus two now because i'm gonna get a common denominator here i'm just gonna write that all over my double bracket which is x plus one and three x minus two there we go now we can go about expanding all of that and just get it all tied up so if i expand the top we get eight x plus eight and then we get plus 18x minus 12 and that's going to be all over this double bracket expanded so x times 3x is 3x squared i'm going to skip a step here i'm going to do the minus 2x plus the 3x minus 2 plus 3 gives me plus x and then minus two at the end there we go that all equals two before i set it equal to two i'm just gonna tidy up what's on the top there we've got eight x and eighteen x which is 26 x and plus eight minus twelve which is minus 4 and that's all over 3x squared plus x minus 2. so now i've got what single fraction with a common denominator and that equals 2. now we can go about actually getting this into a single line equation and removing this denominator so i'm going to times both sides by this denominator here i'm going to times that over and that's going to put it next to this 2. so it's going to be 2 lots of 3x squared plus x minus 2. so if i rewrite this now as a single line let's bring it up here we get 26x minus 4 equaling let's expand this bracket here 6x squared plus 2x minus 4. and if we rearrange this i'm going to make it equal 0 on the left so i'm going to minus 26x and i'm going to add 4 to both sides so i'm going to minus 26x and add 4 and that's going to allow me to get 0 over here obviously it's a quadratic and we need it to equal 0. doesn't matter it's on the left and right but i want to avoid having a negative 6x squared and having an extra step there so 0 equals 6x squared take away the 26x from 2 gives us negative 24x and then add 4 to that minus 4 gives us a zero so there we go doesn't have a number at the end so we can actually solve this now so i can solve this but it doesn't have a number at the end so it's not going to go into a double bracket which it might do if it has a number at the end so we might have two solutions like in that way but i'm going to see what solutions i get this way so if we factorize this to start with um i could actually simplify i could divide both sides by six and if i do that let's see what we get divide both sides by six and we get zero equals x squared minus four x there we go now i can factorize it as well i've just saved myself a bit of time there so if i factorize it we get x equals sorry zero equals x brackets x minus four there we go so now it's factorized hopefully remember as well we've done this on a couple of other questions this x value here has no number with it so one of our solutions is x equals zero and this one here we flip the sign so x equals positive four and there's my two solutions x is zero and x is four and there we go that's that's solved obviously lots of different algebraic fractions that you could have you could be times in and dividing them adding or subtracting them solving with them okay always just look to follow normal fraction rules and obviously looking to factorize where possible in these questions okay so we've got a functions question so we've got f of x is 2x minus four and g of x is x squared plus five now the first one we're gonna have a look at is this here so working out an inverse function so an inverse function just means basically doing it in reverse so the function of f is two x minus four it times this number by two and then takes away four there's a nice way of doing this algebraically we can change that f of x piece there for a letter i'm gonna just change it for the letter y so y equals 2x minus 4. i'm just going to rearrange it to make x a subject which is hopefully quite nice and easy in comparison to the one we looked at earlier so add 4 to both sides you get y plus 4 equals 2x and then divide both sides by two you get y plus four over two equals x there we go i'm just going to rewrite that with x instead of y and that'll be the reverse function there the inverse function so instead of being y plus four it's x plus four so x plus 4 over 2 and that equals the inverse function of f so f minus 1 x there you go and that's how you do your inverse function and that can be true that can be done for any any type of function here just swap the x and y rearrange it to make it as a subject and then rewrite it okay on to the next one it says workout gfx so you could be asked to put a number in this is actually asking us to put a number in it saying gfx which means i have to sub one of the functions into the other so if you read this it means what is the function of g when you put f into it so what is g when you put f into it so well g is up here so we've got to do is put f into it so if i stick f into there and all the way that i do is i stick the function of f in a bracket and i'm just going to replace the x value here with that bracket and if i do that i get 2x minus 4 in brackets squared because it's x squared and then plus 5. so really there is an expression for that but i'm going to completely expand it and simplify it all down so we have a double bracket we have two x minus four and another two x minus four and then once we've expanded that we just need to remember to add five so if we expand that and again i'm going to skip some steps here i'm going to 2x times 2x which is 4x squared and then we've got 2x times -4 and 2x times -4 again so minus 8x minus 8x is minus 16x and then 4 times 4 gives us 16 but we're going to add 5 to that so that will give us 21 so plus 21 and there's your expression for that you could have had it had this written in a different way you could have had f gx and in that case you'd do it the other way around it says what's the function of f when you put g into it so you put your bracket around g and slot it into the x place in f so you can have two different ones there as well i just got to make sure that you do them in the right order you could also be asked to sub a number in so it might say something along the lines of f let's do it the other way around f g i don't know two and in that case that means what's f when g is two so you put two into g and if you put two into g you get two squared plus five which is nine and then you put nine into f so what is f when g is two well when g is two we get the answer nine so you stick nine into f so putting nine into f which is our first function there you do two times nine or two lots of nine take away four two lots of nine is eighteen take away four is fourteen so that would be our answer there if it asked us to sub a number in but these ones are a bit harder when you have to put a function in and uh remember what order to do it in but just remember the way you read it so for this one we looked at what is g when you put f the f function in and there's some functions for you okay so looking at some iterations now this says using x to the n plus one equals five over x n squared plus two and then it gives us x naught equals negative 2.5 find the values of x1 x2 x3 so all this is a process to finding estimates to a solution of an equation now the equation's not given to you in this one but it does give us this iteration formula now this little n plus one here just means to find the next estimate so if you read that in words it says to find the next estimate you do five divided by and this little x n here means the current estimate so the current the current estimated solution squared plus two and then it says when you with x naught equals zero and x naught just means your starting number and it says start with negative two point five this is quite nice and easy actually when you do these all you've got to do is sub that number into the equation so don't be put off by the language here that just means to find the next number so to find the next number all we're going to do is in our calculator we're going to do 5 divided by and this will give us x1 our first solution that's what x1 is 5 divided by the current number squared so negative 2.5 squared and again you've got to be super careful here because you've got to put this in a bracket so the current number squared so when we're putting a negative number in we want to put it in a bracket and then plus 2. so to find the current number to find the you know the current solution or the next solution you do five divided by our starting number squared and add two so literally just typing that into your calculator so open up your fraction button make it look exactly the same as that five on the top negative two point five on the bottom squared and then come out here fraction and plus 2 or press equals and then add 2. now you get a decimal for this you get either a fraction 14 over 5 or 2.8 so there is our value of x1 now obviously now that is our current solution so 2.8 is our new number so i'm going to repeat the process here but instead of putting negative 2.5 in i'm going to put this 2.8 in so where i put it in here that number is now just going to be 2.8 and it's quite a nice little trick you can do on the calculator here as long as you don't press equals you can do 5 over and you can either type this number in 2.8 squared or you can just press your answer button so there's an answer button next to your equal sign usually just says ans and you can just put the answer in there and it's a lot quicker if we have to do lots of solutions here but for this one it's quite nice and easy for me to type it in so i'm just going to write it there we go so you go back on your calculator into your formula make sure you haven't deleted anything press your answer button and it'll put 2.8 in for you press equals again and again this is where the answer button comes in because the number that we get here is absolutely horrible and i'm gonna have to write it all down so it gives me the answer two point six three seven seven five five one zero two and that is my value of x two now i don't really want to write that in over and over again you can do as long as you've written it all down but because i can use this answer button now i'm just going to not touch my calculator screen at all and i'm just going to write down what x3 is so on my calculator screen it currently says 5 over answer squared plus 2. so all i'm going to do is just very carefully making sure you haven't deleted anything off your calculator i'm just going to press equals again and it gives me my next answer it subs that 2.63 number in for me and i get the answer 2.718622914 and it doesn't say to round them so i write down all the numbers on my calculator display and there they are there's the values of x1 x2 and x3 basically just subbing a number in getting an answer and then subbing your answer back in so remember these are estimates to the solutions of a particular equation now the equation wasn't giving us given to us but these are the estimates to the solution of that equation okay so that's what we use iteration for and there's your iteration okay so we've got and i can just see down here we've got some graph transformations so it says the coordinates of the minimum point of a curve are three and minus five now there would be an equation for this curve it's not written down on this question okay and it says write down the coordinates of the minimum point with these equations so given that that line that curve there has an equation what we're going to do to the function over here so there's a few rules to remember with this one if the number in these functions over here is next to the x or whether it is outside of the x and it's next to the y let's say there we go so if it's next to the x it does the opposite of what you'd expect it to do we'll talk about that in a sec okay and if it's next to the y it does what you expect we'll talk about what you'd expect in a second okay if it's next to the x it affects the x-coordinate and if it's next to the y effects the y coordinate well let's have a look at this to start with them so part a there we've got a plus 2 next to the x so it's going to affect the x coordinate but it's going to do the opposite of what we'd expect now we would expect a plus two to add two to the x coordinate but because it's in the bracket there just like when we you can cycle sort of almost link this back to when we're solving and factorizing and when you get a bracket and you flip the symbol it does the opposite we can kind of almost link it to that but what it does is it actually takes two away from the x coordinate so that's gonna mean it's gonna be not three but one so one minus five there you go it does the opposite of what you'd expect we take away two from the x coordinate now on to the next one and this is where this kind of um has a little bit of difference going on um we've got a minus there now it's not next to the x-coordinate so we say it's next to the y so as it's not next to the x-coordinate inside the bracket um it it affects the y coordinate but a negative symbol there just flips the flips the um flips the sign in front of the number it doesn't do the opposite or not do the opposite of what they expect with the negative symbol it literally just changes it from negative to positive or positive to negative so it doesn't affect the x coordinate because it's not in the x bracket so it stays as a three but the y coordinate goes to positive five there you go so it changes it from negative five to positive five okay so the negative could be inside the bracket if it was it would change the symbol on the x coordinate instead look at the next one here we've got a minus three at the end nice not inside the bracket so it does what you expect and we would expect minus three to subtract 3 from the y-coordinate so it doesn't affect the x-coordinate but it does subtract 3 from the y-coordinate so minus 5 would go down to minus 8. there we go every time i'm just referring back to this coordinate here and on to the last one we've got 2x in the bracket okay for this one so it does affect x because it's inside the bracket next to the x but that means it does the opposite of what you'd expect and that means 2 times x so you would expect that to double the x coordinate but if it does the opposite it halves the x coordinate that's the opposite of times n by 2 is dividing by 2. so instead of it being 3 it would be 1.5 and it doesn't affect the y coordinate except b minus 5. there you go it's obviously lots of different variations of this you could be asked to draw the graph as well but you can always just take this approach pick the bigger coordinate on the graph and see how it actually changes the coordinate so it might be that you actually have to move it right move it left move it up or down stretch it or or reflect it with one of these white one of these negative parts outside or inside the bracket okay but always just remember these little rules so if it is in the bracket next to x it does the opposite of what you'd expect and affects the x coordinate and if it's outside of the bracket which i put next to y there it does exactly what you'd expect but it affects the y-coordinate okay so obviously different variations of these if we change our pluses and minuses on some algebraic proof it says prove that the sum of the squares of two consecutive odd numbers is always two more than multiple of eight there's loads of different questions on algebraic proof and again i'll have a video of lots of these different examples but there's just a quite a good one here so it says prove that the sum of the squares have two consecutive odd numbers so there's a few numbers that you need to know so any number we refer to as n an even number well any an even number if you get any number and times it by two it becomes even so an even number in this proof is we're going to say is 2n now to make a number odd we can just think well okay well times it by 2 makes it even and then add 1 to it so an odd number is two n plus one okay so times any number by two you get an even number add one to it and it becomes odd so two n plus one is our little expression we're gonna use for an odd number but it says the sum of the squares so that means add them together and once they've been squared and it's two consecutive odd numbers well i've got one odd number there two n plus one so my next odd number would be two n plus three okay to get to the next odd number we'd add another two to it so it'd be plus three instead so the numbers that i've got i've got two n plus one and i've got two n plus three and it says the sum of the squares which means we've got to square them so i'll square them both and the sum means to add so we're going to add them both together and again lots of different variations we could have here but we've just got to square these so if i square this first one that's a double bracket remember so we'd get four n squared plus we'd get two n and two n so four n plus one and for the one on the right here we'd get four n squared again but it'd be plus six n plus six ends that would be plus twelve n and then three times three is nine so plus nine all right let's have a look what happens when we tidy this all up so adding together the four n squareds gives us eight n squared four n plus twelve and is 16 n and then one plus nine is ten now what i could do at this point is if it said prove that it's always even i'll prove that it's a multiple of two i could i could just factorize it by two and that would show that any number that i sub in there would be multiplied by two but it doesn't say that it says it's always two more than a multiple of eight so i'm looking to try and factorize this by eight and if you have a look while the first piece divides by eight doesn't it i can do that let's have a look divide that by eight or factorize away we get n squared the next one we can divide by eight as well we get two n 16 divided by eight is two but i can't divide this last piece by eight eight does fit in once though doesn't it but it leaves us a remainder and my remainder would be two so if i think about that over here if i split this number up and you can do it like this or think of it as a remainder but if i was to write plus eight plus two instead of plus 10 the 8 would divide by 8 giving me this plus 1 and then i would still have this plus 2 at the end that i've not factorized and that there proves that it proves that it's always 2 more than a multiple of 8. if we subbed any number into this bracket here so any number it would be multiplied by eight and then at the very end we would add two so that there proves it okay we would multiply everything by eight meaning it's a multiple of eight that means multiplied by eight and then at the end we would add two so that proves that any number we sub in there into this expression is always going to be two more than a multiple of eight because we times b and then add two at the end and that's the end of that one there again lots of examples on this that i've got a good video on on some coordinate geometry we've got a straight line passes through the coordinates naught 1 and 10 6. find the equation of the line perpendicular to a b that passes through b so in order to find a perpendicular line we need to find what the gradient is first so if i have a look at a and b we've got naught 1 and 10 6. so to find the gradient we can do the change in y over the change in x which you might also refer to as rise over run so the change in y goes from one to six so the change in y there is five from one to six the change in x is from naught to ten so from not to ten that's ten and if we simplify that that gives us a half i always leave these gradients in fractional form rather than as a decimal so the gradient is a half and i always write m equals a half because we know we're looking at line equations which is your y equals mx plus c so m is a half it says find the equation of the line perpendicular to that that passes through b so we're using this equation so a perpendicular line we do the negative reciprocal so to get my perpendicular line which i'll call mp you flip the fraction over to get the reciprocal so two over one and you change the sign so it's negative two over one and negative two divided by one is just negative two so the gradient that we're looking at here is negative two so what i'm gonna do is i'm gonna write out what the equation of a line always looks like which is y equals mx plus c and i'm going to plug this piece into the equation of my line because i've now got the gradient so we've got y equals negative 2x plus c now it says it's got to pass through that point 10 6. so it's got to pass through 10 6. so i'm just going to stick these numbers in remembering that we have an x coordinate here and a y coordinate here so if i stick those in we have y is 6 so 6 equals negative 2 multiplied by the x coordinate 10 plus c now we just need to solve this for c so we have 6 equals multiply these negative 20 plus c add the 20 over and you get 26 equals c so you found your y-intercept there all you've got to do is finish this question off put all the pieces back together so you have y equals minus 2x which we have that had already had up here but now we've got c and that's positive 26 so plus 26. so y equals negative 2x plus 26. again you could rewrite that in another way you could write y equals 26 minus 2x i'm just going to leave it like that because it's fine to write either way and that's the equation of that perpendicular line okay the last one here we've got a circle a has the center two five and the point p eleven eight lies on the circumference of the circle find the equation of the tangent to the circle now what we know about tangents is they are perpendicular to the radius okay so a tangent is perpendicular to the radius let's think about that perpendicular to radius now it can help with these types of questions to draw a little bit of a diagram just have a think about what this actually might look like so if i have a think a bit of an axis here it says the center is at 2 8 so i'll draw a little sketch not the best sketch here but 2 8 and it says the circle point p is 11 8 let's imagine that's over here somewhere so we've got a tangent coming off the circle and that's what it looks like so there's a few that's a really bad center there let's change that it's a little bit better there we go so this is a really really general sketch of what it might look like okay and it says the point uh center has a coordinate 2 5 which is there and point p is up here 11 8. there's lots of information we can find on this question here but let's just have a look and just see what it's actually asking it says find the equation of the tangent so in order to get the equation of the tangent i need to get i need to know what the gradient of it is if we think about the radius here okay bearing in mind that they're perpendicular that's what we've already stated it forms the right angle could we find the gradient of the radius now we've got these two points so we've got two five and that definitely attaches to 11 8 which is on the circumference so i can find the gradient of those points i can do the same as we did in the last question i can do the change in y over the change in x and that's change in y goes from 5 to 8 which is 3 and the change in x goes from 2 to 11 which is 9 so my gradient is a third there we go so my gradient or m is a third so looking at the tangent which is perpendicular i can say the gradient of that is well flip it over would give us 3 change the sign negative three so there's the gradient of the tangent so if we stick that into our line equation as the tangent is a straight line we get y equals minus three x plus c and we know that it passes through this point eleven eight okay otherwise it wouldn't be on the circumference of the circle there which it says it is that's where the tangent meets it so what i'm going to do is i'm going to stick 11 8 into the equation of the line just like i did on the last question so we get 8 is y equals negative 3 multiplied by 11 plus c and again exactly the same as the last question we can solve this now so 8 equals negative 33 plus c times in those out add the 33 over 41 equals c so now i've got the y intercept of that line as well now if we put this all together we've got our equation so y equals negative 3x plus 41 there we go and there's the equation of the tangent a few other things that you could be asked about this question though um let's have a look there we go you could be asked to work out maybe the length of the radius so if we go back to our diagram over here in order to get the length of the radius we'd have to use a little bit of pythagoras so bringing that down there you've got a little right angle triangle you can draw in and it goes from 2 to 11 which was a run of 9 there and a rise of three so you could do three squared plus nine squared so you could actually get the length of the radius there what would that be three squared is nine nine squared is eighty-one so square root of that is root ninety so you don't need it for this question but you could say the radius is root ninety you could be also be asked to write down the maybe the equation of the circle or something like that um but just something else to be thinking about if you ever want to find the length of the radius and you can you can find it by using a bit of pythagoras there not that we needed it for this question though right okay that's the end of algebra okay so moving on to some geometry so we're going to start off with this volume of a frustum now you would be given the volume of a cone formula and that is volume equals one-third pi r squared h where r is the radius and h is the height so if we have a look at this for a stem here the only thing it doesn't give us is this length across the top but if we have a look the height of the little cone on top is 20 and the height of the big cone if it was still there is 40. so it's half the size of the diameter there is 15. now in order to work out the volume of a frustum you got to work out the volume of the big cone if it was there and we're going to take away the little cone that's been chopped off the top so i'm going to start off with the big one i'm just going to label this the big cone and i'm just going to plug the numbers into the into the into my calculator so it's 1 3 times by pi times by the radius squared now careful because it's not 30 it's 15 half of that so 15 squared times by the height which is 40. so typing that one into the calculator and i can leave that in terms of pi or i can write out the full decimal and it comes out as nine four two four point seven seven seven nine six one centimeters cubed now for the small cone on top there we go we can work out the volume of that as well so it's one-third times pi times the radius squared careful it's half of 15 so 7.5 squared and times up by the height of that cone which is 20 and that gives us a volume and again if we just type that in let's have a look so one third multiplied by pi multiplied by 7.5 squared times 20 and that gives us a volume again i could leave it in terms of pi but one one seven eight point zero nine seven two four five and again that centimeter cubed so in order to work out the volume of what's left when the little cone has been cut off the top we just need to do the big cone subtract the volume of the little cone there so i'm just going to subtract this away from that so subtract and again i'm just going to do that on the calculator so 9424.7 take away that answer there leaves us with the total volume of 8246 points and i'm going to round it to decimal places although you would be asked how to round in a question i'm going to go for 0.68 centimeter cubed just be careful what the question asks for obviously if it said one decimal place it would be 0.7 it might even just say four or three significant figures in which case you just got to make sure you just round out how the questions asked but there you go that's using the volume of the cone formula on a frustum now which is a little bit more interested in just doing the normal cone next one here we have a hemisphere and it says work out the volume of the hemisphere now again we'd be given the formula to work out the volume of the hemisphere and volume of the hemisphere is four thirds pi r cubed and that is our final volume there oh which makes that says that's a rubbish three let's change that one four thirds pi r cubed so we've got to do is stick the numbers in the formula again so if i want to work out the volume all i have to do is in the calculator i'll do four thirds multiplied by pi multiplied by eight cubed and we'll get our answer here straight away so fraction button four thirds multiplied by pi multiplied by eight to the power of three and we get a volume here of two one four four point six six again if i round it to two decimal places centimeters cubed just remember obviously to write all the numbers down and only round it once you've been asked to okay so once you've got your total volume of your sphere there just remember this is a hemisphere so it's half the size of a sphere exactly half so i just need to divide my answer by two here and once we divide that by two we get a final answer of let's have a look one zero seven two so one thousand seventy two point three three again i've rounded it there centimeters cubed and that would be your final answer here for the volume of the hemisphere not forgetting as well you could be asked to work out the surface area and if you're asked to work out the surface area again it gives you the formula of a sphere and the surface of a sphere is 4 pi r squared so again if i was to work out the surface area of this one it's a little bit more complicated um but yeah only in the sense that we work out the full surface area so we do 4 times pi times 8 squared if i just go for this let's work it out 4 times pi times 8 squared that gives us a surface area of a sphere being 804 and it's 0.25 as well again we'd have to halve it because it's half okay so half of that would give us a surface area of 402 0.12 and that is obviously a surface area so centimeter squared but if you're working out surface area you've got to make sure you don't forget as well the circle sitting on top there because it's the total surface and the area of a circle is pi r squared so we can work out the area of the circle as well if we just do that pi times h squared we get an area of the circle there of 201 0.06 and you would just add these two numbers together to get the total surface area so if you add 201.06 we get a total surface area of 102.12 added to that gives us 603.18 centimeters squared again so i know we were looking at volume there but just thinking about if you had to do the surface area as well how you'd approach that when you've got a hemisphere and you've got to make sure you halve them and add on the extra circle okay so we've got some similar shapes two cones are mathematically similar the height of cone a is four and the height of cone b is ten now you can always imagine what this is gonna look like you can always draw yourself a picture if you haven't been given one and we've got a bigger cone so we've got a height of the smaller one being 4 and a height of the larger one being 10. now straight away you can work out a scale factor between these two if they are mathematically similar it just means one's an enlargement of the other so i can do the bigger length 10 divided by four and it gives us a scale factor of 2.5 now when it comes to scale factors that is our length scale factor and always label it length scale factor now there's two more scale factors we could we could want to look at and that's the area scale vector or the volume scale vector and to get from the length scale factor to the area scale factor you square it okay so we square the scale vector to get from the length scale factor down to the volume scale factor you have to cube it so if we just have a look at this question here it says the volume of cone a is 40 work out the volume of cone b so that's cone a is the smaller one in this case i should probably label that a and b and it says the volume of this one is 40. so in order to find my volume scale factor i need to do 2.5 and cube it to get a volume scale vector so we'll do that on the calculator 2.5 cubed gives me 15.625 and that's my scale factor in terms of the volume so the volume of cone b is going to be 15.625 times bigger so in order to get that volume there we're just going to multiply 40 by 15.62 so times up by 40 and we get a volume here of 625 centimeters cubed there you go just remember as well if we were going from the bigger down to the smaller we'd have divided by the scale factor but as we're going smaller to bigger we're going to multiply so look at a different one so a similar question two cylinders are mathematically similar we've got cylinder a and cylinder b now it's a cylinder a is 160 and cylinder b is 20. so a's my larger one so if i just draw a little basic diagram of this we've got a and we've got b there we go and this time it says it actually gives me the volumes so it says the volume of a is 160. the volume of b is 20. so again if i find a scale factor here i can do the bigger one divided by the smaller one 160 divided by 20 gives me a scale factor of 8. now thinking about this area scale factor length scale factor and volume scale factor there we go i've got this time i've got the volume scale vector and that is eight but i can't get from volume straight to area now if you remember to get from length to volume we had to cube it okay so to get back from volume to length we have to cube root it so if i do the cube root of eight that tells me the length scale factor is two now the question here is saying it then starts it asks me to work out the surface area so i need the area scale vector and if you remember just from the last one to get from area it's length down to area we square it so that'll be 2 squared which would be 4. so my area scale factor is four okay so whenever you're given an area or a volume you have to go back to length first to then get back to an area or a volume so my scale factors four for the area and it says the surface area of cylinder a is forty so that's the bigger one i'll just write area rather than surface area so to get from the bigger one down to the smaller one we're going to have to divide by that scale factor which is 4 for our area and 40 divided by 4 gives us 10 so we get a surface area of 10 centimeters squared all right there we go there's thinking about some similar shapes okay so we've got a transformation here it says in large shape p with a scale factor of negative a half with the center of enlargement zero zero so always mark out your center of enlargement first and an enlargement with a negative scale factor and a fraction here means we've just got to do this very very carefully now the first things first you need to pick a point on the shape and just figure out how you get from the center of enlargement to that point there and if i just count that that's one two to the right and one two three up now what i need to do to do this and it's going to be quite small in this diagram is i just count in the opposite direction by whatever the scale factor is that's what the negative is so rather than going from from zero zero rather than going right and up i'm going to go left and down but it's a scale factor of a half so i need to also halve those distances just as if it was negative two i would double those distances but for negative a half so i'm gonna go left one rather than two and i'm going to go down rather than going up three i need to go down 1.5 or one and a half so one and a half gets me to there there we go so that's one and a half i'm gonna do the next one in a different color so i'm gonna pick this point here and let's just think how we get to that let's get rid of some of these markings here there we go so to get to that we go one two three four across and one two three up so i'm just gonna half that again i'm gonna go two left rather than four right so one two and then that one and a half down which again gets me down to there there we go and that is that point then again i just need to repeat the process for the last one you might be able to start to do these in your head once you've had some practice but if we do the last point here being the top one let's just see how we get there so it's one two across just like it was to that red one and then we gotta go up one two three four five six seven so half of that's going to be 3.5 so i'm going to go one to the left okay so i need to go down 3.5 1 two three and a half just there there you go obviously you can do all this in pencil to keep this all nice and tidy but when you've got a negative scale factor you're just going to go backwards in the opposite direction by whatever that scale factor is obviously join it all up nice and neat and there we go and that is that one there just always go back and double check so it was seven up so we went 3.5 down that's absolutely fine the other one was three up and we went 1.5 down half of that and the other one there perfect okay so you can see with these it does actually get bigger and it gets bigger or smaller and it rotates 180 degrees and that's how a lot of negative enlargement does to a shape okay so we're going to have a look at another one here okay so it says describe the single transformation that map shape b onto shape a this is quite nice you hopefully you can tell that it's an enlargement okay because obviously one's got bigger than some bigger and smaller and we're going from b to a now what you can do is you can get your ruler and i've got to think i've got to what you're going to have to draw and do this quite carefully but you get your ruler and you join up the similar points now the similar points are here and here okay obviously it's been rotated 180 degrees now if you get your ruler and join that up you end up with a little line like that and then you just do that for all the other points so i can probably get away with just doing two we'll have to wait and see so if i join up these two similar points here there we go they join up just like that and what you'll find is there's this little crossover point and that tells you where the center of enlargement is so i know that it's an enlargement we've already got that so i'm going to have to state it's an enlargement okay so it's an enlargement i've got to get what the scale factor is i haven't got that quite yet but i'll have a look we know it's negative as it's getting this uh 180 flip so you know it's negative we'll come up with a number in a sec and i know where the center of enlargement is so center of enlargement we just write down what that coordinate is and it's it's two two so center enlargement two two now to count that scale factor we're just going to count this side here maybe and this side here and just see how much bigger it's got so that's gone from a length of one to a length of three so that is a negative three scale factor so it's an enlargement scale factor negative three center of enlargement two two i can use that approach for any of these enlargements whether it's negative or not also you might just wanna note here that if it had been from a to b it would have been the other way around it'd have gone from three down to one and my scale factor instead if i was going from a to b it'd be exactly the same description there but my scale factor would be negative a third because it's getting a third of the size going the other way okay so two different ones but the description would be exactly the same there okay so we're gonna look at some circle theorems for the purpose of keeping this video as short as we can i'm only going to go over a couple of them and just how to approach the question if you want to have a look at every single circle theorem how to tackle every type of question you have to look at my circle theorems video now some information to go along with this i'm going to say that ap and bp are tangents there we go so when it comes to tangents they meet at equal length that's one of our circle theorems so these two lines here are equal length meaning that a p b that triangle there is an isosceles meaning that the base angles these two are the same so there we go we can start to figure out some of the angles here because if we do 180 the angles in the triangle take away 86 it leaves us with 94. and that we can split in half to share it between our two triangles there so 94 divided by 2 is 47 so both the angles at the bottom of the triangle there are 47 there we go and as with all these questions it always says to state your reasons so you would say tangents meter equal length therefore this triangle is an isosceles okay and writing that down okay on to the next bit we've also got um and it doesn't doesn't say it here but o is the center of the circle now o to b there is a radius looking at this line just here and a radius meets a tangent at 90 degrees meaning you can always draw this on the diagram when you see a tangent meter in a radius we've got a right angle just here so we can work out the value of x because if the full angle is 90 we gonna do 90 take away the 47 and that leaves us with 43 degrees the important part with this question is to make sure you write down all those reasons so we would write the tangents meter equal length isosceles base angles are the same and then our final reason for this bit of the working out the tangent meets the radius at 90 degrees and therefore we could do 90 take away 47. okay a different circle theorem question with some different circle theorems within this one so if you have a look we've got these points a d c b around the circle and that forms a quadrilateral and these are called the cyclic quadrilaterals and the rule here is the opposite angles add up to 180 so we've got the 70 over there so opposite that is the nine is the y and opposite angles have to add up to 180 so we could do 180 take away 70 and it gives us our answer of 110 degrees and our reason for that which we would have to write down again is the opposite angles in a cyclic quadrilateral add up to 180. the next one here is to find this angle x and that's going to involve one of our other circle theorems and when you've got these points made at the center i'm just having a look at points d and b i always do this with a little highlighter but if i make this angle 70 here i can also from the same two points make this angle here at the center and that's one of our other circle theorems angles at the center are double angles made at the circumference from the same two points so to work out angle x and this first one we did was y to work out angle x we would just do 70 times two double it and that would give us 140 degrees and again that would be accompanied by the reason angles at the center are double angles at the circumference when they're made from the same arc so our two answers here are 110 degrees and 140. again just a few little bits of circle theorems there to be having a thinking about and making sure that you are writing down all those reasons that's absolutely crucial for these okay so we're on to some congruent triangles it says prove angle abd is congruent to cbd now if you have a look abd is the triangle on the left and cbd is the triangle on the right now in order to prove triangles are congruent we need to look for similar sides and similar angles now the first one that pops into my head is the line b to d this line in the middle okay so i highlight that this line b to d in the middle is the same for both triangles so i could prove a side straight away so i can say aside i've got b d is the same in both so both triangles already share a similar side or exact same side bd is the same in both remember we're trying to prove that they're congruent which means that they're exactly the same so let's get rid of that we can have a look at another one now can we see any other sides that are the same well we've got these little symbols and i've already got an arrow pointing to two of them really we've got these little symbols on all of them which means that all of those lengths are the same length so we could state any of these are the same now we've got a b this one here number one is the same as bc number two so i can do another side so i can say that a b equals bc and actually that's given to us in the question so i would just have to say given in question there we go and actually using that same logic there we could actually prove that these two down the bottom which are labeled three and four are the same as well because they're all the same length so i can prove another side so i could say that ad equals cd and again that's given in the question there we go so what i have to do is state why that they're congruent there and uh i can just say and i'll just write this down a box too i can say therefore they're congruent okay because the because the the sss rule here so side side side congruent and we're using the sss rule okay so there's other things that you can look for as well you can have a look for angles obviously there's a few of these that you can use and it's just an idea of how to approach it and how to lay it out just labeling what you found but because these are obviously isosceles triangles so this one on the left here we've got two sides at the same length so we know that the base angles are the same here therefore as they're all the same length those base angles must also be the same as the base angles on the other triangle so we could use that as well and if those angles are the same then this angle of a has to be the same as the angle at c so actually we could prove all of them so there's a lot of different types of questions here a lot of different proofs but there's just an idea of one of them and how to lay it all out okay so we're gonna have to work out some missing lengths in this triangle now it's not a right angle triangles to find missing lengths and angles we can use either the sine rule or the cosine rule and these are rules that you're going to have to remember but we'll have a look at one to start with and how we know when to use it now the first thing you do is you try and identify in the triangle first of all what we're looking for which is a b so let's label that x now what i look for straight away is do i have pairs of opposites so i have this pair of opposites here and i've got both of those and then i've also got x opposite to this angle so i don't have x but it's in one of my pairs of opposites so when we've got this scenario where there's two pairs of opposites we can use the sine rule and we only ever need part of the sine rule so i'm just going to use a over sine a equals b over sine b also equals c over sine c but we only ever use i only ever need to use two of them here so let's have a look we need to label this up and i'm going to completely ignore the letters that are on the actual triangle itself i'm going to label this angle a which it already is and this and this side little a opposite that and then this angle b and the one opposite little b and all i need to do now is stick all the numbers into the formula so let's have a look a is 12 so it's air 12 over sine the one opposite that 55 is equal to b which is our x over sine b obviously you should already know this two for two variations of this formula we could have it flipped over so we could have sine a over a equals sine b over b but this is our one for side length and we know that because our unknown piece is on the top so we're able to isolate this now quite easily so sorry i've written b there it should be 20. there we go sorry there we are 20. so what we need to do is times both sides now by sine 20 what you could do is work this out on your calculator and times your answer by sign 20 but i'm going to multiply it straight over so i can type it all in one go so times by sine 20 and if we do that we get 12 sine 20 it goes on to the top there over sine 55 there we go and all we have to do is type that into your calculator obviously just being careful that you put these angles in brackets some calculators are going to need you to do that so if we type that all into the calculator again not forgetting you could just work out 12 over sine 55 first and then times it by sine 20 but i'm just going to go for it like this so 12 sine 20 on the top closing your bracket over sine 55 and on the calculator just writing down what you got you got 5.01 one zero three five three nine nine eight now a question would normally say how to round this so if we imagine it was two decimal places for this one it would be five point zero one and it's a length so centimeters okay so just obviously just be careful the question says let's have a look at an idea where we've got to find the angle okay so in this question let's have a look work out the size of angle bac so let's identify that that is here bac okay so we're going to use the formula the other way around this time so sine a over a equals sine b that does not say thing b sine b over b okay so plugging in our numbers let's just label it up so let's call this one a as the a is next to it that's fine and again i'm just going to write over this one i'm just going to put b and b okay just because the way i've written my formula so then sticking in all the numbers what have we got sine a is sine x so we have sine x over 20 equaling sine 43 over 14. okay so exactly the same approach as we did before we can isolate the sine x by timesing both sides by 20. and again you could work that right hand side out and times it by 20 but i'm just going to go stick it up the top there so we end up with sine x equals 20 sine 43 over 14. if we type that into the calculator now what do we get 20 sine 43 over 14 and we get an answer here let's write it over here so we get sine x equals 0.97428 and a few more decimals and obviously just like normal trigonometry when you're doing socatoa to get the actual x here we have to the inverse of sine so we're leaving that number on your calculator you do shift sine which gets you sine minus 1. type in that answer or just press your answer button so shift sign answer press equals and i get an answer here of 76.9779 and again a question what else is to round it here just depends so let's just go to the nearest degree we go for 77 degrees obviously just making sure what the question says here but let's just round it to one the nearest degree there so 77 okay so that's how you use the sign rule right let's see how this question is different then so work out the length of a b so this one over here now straight away looking at this look we've got a pair of opposites there but i don't have any other pairs of opposite opposites i've not got anything opposite my 15 i've not got anything opposite the 12. so i can't actually use the sign rule here and that is your clue that is your hint here that the sine rule is not going to work we're going to have to use the cosine rule so another rule that you need to know so the cosine rule is e squared equals b squared plus c squared minus 2 bc cos a okay so a being the side we're looking for so we'll label this little a and this one big a ignoring the letters on the triangle and then labeling the other two sides and they can be b and c however you like and from there all you've got to do is stick these numbers in it's actually quite simple to use when you know it so a squared equals b squared which is 15 squared plus c squared which is 12 squared minus and i'm going to stick this bit in brackets 2 times 15 times 12 cos a which is down here which is 20. there we go right so sticking that all in the calculator let's have a look what we get so 15 squared plus 12 squared minus 2 times 15 times 12 cos 20. press equals and i get a squared equals 30.7106 and a few more decimals now obviously that is a squared and we don't want a squared we wanna know what a is so we just need to square root both sides now so square root leaving the answer on the calculator square root answer and we get a equals 5.541719 and again obviously you'd be asked to round this in a particular way in the question let's go for two decimal places so a equals five point five four centimeters all right there we go and there's using a bit of the cosine rule okay so working out the size of angle bac which again is this one at the top and again just having a look there are definitely no opposites because we've got no other angles but uh the angle that we're looking for is going to be our a and it is opposite nine there we go so the others can be b and c again now obviously here we're looking for an angle so it's up to you if you choose to learn the formula i tend to find that i just learn this formula a squared equals b squared plus c squared minus 2 b c cos a and then i'm quite happy just rearranging that to get cos a on its own so in order to do that i'm going to get this whole minus 2bc cos a i'm going to add it to the other side so we get a squared plus 2 bc cos a equals b squared plus c squared now i can get rid of that a squared so we can minus a squared from both sides so minus a squared and you get 2bc cos a equals b squared plus c squared minus a squared now and i can finish off this rearrangement you can divide by 2 bc just to leave you with cos a so cos a equals b squared plus c squared minus a squared over 2bc it's up to you you can choose to just learn that formula if you want but that's the formula we're going to use to find an angle so plug in all these numbers then just into my formula there we'll get cos a equals b squared plus c squared so 10 squared plus 5 squared minus a squared so minus 9 squared all over 2 to 2 times b times c so 2 times 10 times 5. nice and easy typing that into the calculator so fraction button 10 squared plus 5 squared minus 9 squared all over 2 times 10 times 5 and that equals let's have a look so cos a equals 0.44 and then same process again we need to do the inverse of cos so cos minus 1 of your answer and you get let's have a look shift cause answer and i get 63.896 there we go degrees and again we could round that so we could just say 63 point and let's just go to one decimal place 63.9 degrees again just reading the question and that's how to use the cosine rule for finding angles moving on to the area of the triangle so your area of a triangle formula obviously not half base times height because we don't have the height here so for any triangle the area is half a b sine c so half a b sine c another formula there that you need to know and let's just have a look at how we apply that so sine c this time the angle that we're going to use is going to be our c so we'll call that c this would then be little c and the others will be a and b and they can be a and b in either order and then it's as simple as just sticking those numbers into a formula so it's half times times 8 times sine 41 there you go type that into the calculator so half times 5 times 8 times sine 41 and we get the answer 13.12 if i rounded to two decimal places it's area so that's meters squared just being careful of the units so 13.12 meters squared there we go there's using the area triangle formula question like this though you might be given the area of the triangle so it says the area is 80 so we're working backwards a little bit here so we're going to use half a b sine c again so half a b sine c but this time it gives us the answer so the answer to that half a b sine c has to equal 80. so let's just plug in all the numbers here again c being my angle and then a and b being my other lengths so if i was to type this into the formula we'd have a half times 11 times 16. times sine c and it would equal 80. now i can't actually work that out but what i can do is i can work this bit out so if i do a half times 11 times 16 we get the answer 88 so we have 88 times sine c equals 80. so we can rearrange this look we can get both sides and divide them by 88 and that will just give us sign c on its own so sine c equals an 80 divided by 88 is let's leave it as a fraction actually it is a decimal there but it comes out as not 0.90 and that is actually nine zero recurring there we go i'm just going to leave it on the calculator screen and then obviously to finish that off you do the inverse sign again just like we have on these other questions so i'm going to do that over here to the left so sine minus 1 of this number here which is actually also a fraction it's 10 over 11. so minus 1 10 over 11 or that 0.90 recurring and we get the answer of 65.38 degrees there we go so we can find an angle as well just working backwards if we're given the answer and you can apply that logic there to lots of different questions where it gives you the answer and you can use your formula backwards just dividing everything over to the other side okay so we've got a vectors question it says b is the midpoint of ac so that's this point here is the midpoint so straight away that tells us look because you can see some of the vectors on here that this must also be a and again there are so many different variations of vectors here this is just one idea of how to approach it there's so many different types of questions you can get we could spend an hour to two hours just looking at vectors m is the midpoint of p b so we've got a midpoint here so that's halfway along and then show that nmc is a straight line so straight away i'd like to draw nmc in and just think about that so that it's saying is show that that's a straight line i'm going to have a look at some of the vectors on this line and see if we can get a common multiple or a common bracket however you've looked at this before so all i want to have a look to start with is that line n to c i've got three vectors i can look at i've got the full line from n to c we've got from n just to m so i'm going from n to n or i could go from m to c and they're all little parts of this line okay the full line there let's just highlight this the full line the first one that i've written is n to c all the way the second one i've drawn is n to m from there to there and the third one i can have a look at is from m to c from there to there now it's up to you which one you choose sometimes some of them are easier than others but the first one there so this n to c is quite simple for us to do now to get from n to c i've got to move backwards up the line this way and then i can go forwards along to c there so that first part of the journey going from let's get rid of that going from n up to a and let's draw this in to go from n to a so i'm just going to write this down is minus 2 b you have to go backwards through that 2b so from n to c first of all it's minus 2b then i can go from a over to c doing two a's and that's positive so plus 2a there we go so i've got minus 2b plus 2a now whenever it comes to these vectors here you always just want to look to factorize them so i'm going to move this out the way i'm just going to see what does that factorize to now i can factorize it by 2 and i would get 2 lots of minus b plus a and obviously we could write that in a slightly different way i could swap the letters around if i want i could write 2 lots of a minus b now if i can prove that one of these other vectors here that i'm looking at below has this bracket a minus b then they must be on the same straight line because therefore they're going in the same direction so let's have a look at one of them so i can either look at n to m or m to c it doesn't really matter i'm just going to go with m to c either one here are going to be just as difficult for us to do but that first one there is nice and easy so always make sure you have a go at finding some of the lines now that m letter here is halfway along pb so i'm going to have to find the vector for p to be and that's going to be my absolutely key one here p to be because if i can't find the full length of the line i'm definitely not going to find halfway along it so if we look at how do we get from p to b well that's actually quite simple you can go from p up to a and that's minus three b's so let's write that down you've got you've got the minus b to get to n and another two to get up to a so minus three b's and then to get from a to b you'd do a plus a moving along that vector there so plus a so that is my vector p to b that's another very good one to always make sure you get that missing line where you've got a halfway point now in order to move halfway along that line i'd have to do half of this vector so to get from p to m or from m to b which is half of that let's go for m to be is that's the quickest way for us to move there m to b we'd have to do half of this vector so to times that by a half i'm just going to write it as a half lots of minus 3b plus a so half of that now if i expand that i can i can actually add stuff to this as well because that only gets us from m to b if i draw that on that gets us from here to here doing half of that line so expand that out minus three times a half is minus three halves so minus three over two b plus half of a so half a and now we can finish that off because now we're at b we can move our final little bit of the line which is to go over here to c and that's adding an additional a in there okay so that's our last little bit plus a and if we add this all together and let's see what we get so we have minus 3 over 2b i'm going to write this over here that's not changing so we've got minus 3 over 2b and then we've got half a add an a and that gives us 1.5 a or another 3 over 2 a there we go 3 over 2a so let's write that down over here we have minus 3 over 2b plus 3 over 2 a so that is our vector there to get from m to c now what i need to do is just factorize this one if you have a look they both divide by 3 over 2. so i can take this 3 over 2 out of the bracket and if we divide them both by three over two we get minus b plus a there we go i rewrote that a different way as well i could write that as three over two brackets a minus b the other way around and there we go we've got that matching bracket there in both of them okay so both share that common vector there that common movement between the between the uh between the points on the diagram n m and c again you could have found n to m as well we could have done the same thing and actually because we found this um half this one just here we can actually look just think if i was moving from n to m so n to m we would do plus b and then half of that vector just there and if we did do that let's just do it over t to the side so this is from n to m not that we didn't need this one we only ever need two so from n to m i could do plus b or just b and then add to that this vector down here that we we did in the green down there so add to that minus 3 over 2b plus half a and if we add that all together there minus 1.5 plus 1 gives us minus a half so we'd have minus a half b plus a half a and again we could factorize that by a half so we get a half minus b plus a and again we get that minus b plus a there which we can write in a different way so we could have done any three of them all three of them eventually give us this same bracket here okay so we have the same multiple it means it's on the same line and there's a bit of vectors for you and the last question here is just involving some circle sectors but also involving a little bit of trigonometry as well it says that this is an arc of a circle sent to zero it wants us to work out this shaded region so if we can work out the full sector we can take away this unshaded triangle so to work out the area of a sector we would have to do pi r squared so not forgetting the area of a circle formula pi r squared and multiply that by whatever fraction of the circle this is no it says 35 degrees over there so it's 35 over 360. so if we plug those numbers in it will give the area of the sector so we've got pi times 80 squared it's in meters multiplied by 35 over 360. if we type that in on the calculator let's see what we get pi times 80 squared times 35 over 360. and we get the answer 1954.7 and it rounds around it's two to four places seven seven obviously be careful the question says again that's meter squared then we can work out the area of the triangle and obviously not forgetting just because it's not just a triangle on its own we can still use our half a b sine c so we can use half a b sine c to get this one and if we plug the numbers in 35 there is our c and because it's a circle sector a and b here are the same because it's the radius of the circle so we can do half times 80 times 80 times sine 35 there we go if we type all that in 0.5 times 80 times 80 times sine 35 we'll get the area of that triangle there which is 1835.44 meter squared and then to finish that off we can do the bigger shape the sector subtract the area of the triangle so 1954.77 take away this 1835.44 and that leaves us with an overall area here let's work it out 1954.77 take away answer and we get 119.325 meters squared and there we go and obviously just rounding that however you've been asked in the question so there's quite a large selection of geometry questions there that we're doing on the geometry okay so we're going to look at some statistics this is going to be quite a short one now we've only got a few questions to have a look at and we're going to start with a box plot so on here we've been given some information about some uh time in seconds that 15 people wait to be served at a garden center and it wants us to draw a box plot for this part a so we've given a list of numbers for those minutes and you just got to make sure they're in order because we need to find a few values we need to find the lowest value the highest value the median and the two quartiles the lower and upper quartile so we can quite easily see that they are in order so we can get the lowest and the highest value from either side and they're going to form the ends of our box plot so just very carefully on your scale here drawing the ends in here and 44 just make sure the scale goes up in ones it does i'm going to do as best as i can here and that's our 44 there now we need to find the median so with 15 people we just want to find where that halfway point is so all you got to do is add one and half of it so 15 plus 1 is 16 divided by 2 and that's going to be our eighth person so we just need to go along the box plot and find where the eighth person is so one two three four five six seven eight so 25 is going to be our median that's quite nice it's halfway between 20 and 30. drawing and nice and neat there to form our median now we just need to find where the quartiles are so on that lower half okay the lower part or the lower quarter here if i try and highlight that we've got these numbers here and there are seven numbers there if you just count them up so one two three four five six seven so again finding the halfway point we can do seven plus one and divide it by two which gives us four so we wanna find where the fourth person is for that lower quartile so five nine eleven fourteen there is our lower quarter there so marking that on the fourteenth person one two three four and then looking at the higher quarter there 27 27 28 30. okay so once you've marked all of those on they've got to go to join up your little box there we go making it like the one that you can see below and then join up these parts here and there's your box plot drawn okay moving on to the bit below and i haven't added the question in here but quite often it says to compare the box plots so it says here the box popular below shows the distribution of times served a different garden center and it would obviously you would usually say to compare them there are two things that you want to compare when you're comparing a box plot one of them is the median so if we have a look our median here is sitting on 25 and down it's sitting here on 21 now i don't have to mention any numbers but i would mention number one i would say that rose or just i'll just call it rose rose has a higher median wait time okay so a higher median weight time okay make sure that you mention some context obviously don't just say it has a higher median what's the story actually about it's about the wait time at a garden center so rose is a higher median wait time the next thing we have a look at is the distance between the quartiles and again i don't have to mention any numbers but i just have to think this is the distance between the quartiles it's called the interquartile range and on the one below here is the distance now hopefully you can quite see there visually that on the greens garden center there is a higher interquartile range meaning that the data is more spread out and you want to mention that word spread out so number two i did my comparison here greens has a higher i'm going to abbreviate it to iqr into quartile range meaning the data is more spread out or meaning the wait times are more spread out there you go more spread out and mention that word there spread there you go some more spread out and that is drawing a box plot and also making a comparison between the two let's move on to our next one okay so the next question here we're going to draw a cumulative frequency graph okay so it says cumulative frequency up here and then it says there's some information about the speed of 100 lorries which it starts to tell us over here now sometimes it might give you an extra table down here to draw them in just take note of the fact with cumulative frequency we are we are doing a cumulative total as we are go up these speeds here in our groups so we start off with two which is shown to us in the table just here and then it says in the next part of the table right here it says look naught to 40. now that is correct it should say not to 40 because that's going right from naught all the way up to 40 and that means the accumulation or the cumulative total of 2 and 9 there and if you add up to a 9 we have 11. then there's 23 so from naught to 60 so add the 23 into that as well we have 34 add the 31 into that we get 65 add the 27 into that add 7 is 72 at 20 is 92 and then add the 8 into there gives us our total of 100. now if it doesn't give you a little table to the side you just have to do it over to the side of your table here so 11 34 65 92 and 100 and just drawing it in there instead but sometimes it might give you the table from there it's quite nice and simple we just have to actually draw it now when we draw a cumulative frequency graph we plot these numbers these cumulative frequency numbers against these values here the endpoints okay so depending on which table you've got we're just going to plot it against that so if i do this nice and good as quick as i can anyway because it's quite small but 20 to is here make sure it's nice and accurate mine's not going to be perfect and then we've got 11 the cumulative frequency going to 40. so 11 is here then we have 34 and it just goes along 20 every time so 34 it's around about here then 65 there we go then we've got 92 just here and then finishing it up on the 100 there we go now when it comes to a cumulative frequency graph it is a nice smooth curve just have a look what your starting number is now your starting number if i highlight it is over here it starts on zero so we're going to start from zero and we're going to do a nice smooth curve making sure you go through all the points and there we go there's your cumulative frequency graph now there's a few things that you can read off a cumulative frequency gap graph the actual question that we're going to have a look at down here is find an estimate for the number of lorries with a speed of more than 90 but there's lots of things you can find we'll start with this one so a speed of more than 90 if we have a look well a speed of 90 is here so all i have to do is very carefully go up to my graph and show this on your question and then you go across to the left now this is going to vary depending on different people's curves i'm going to try and read it as best as i can mine looks like it's coming out at around 84. so the answer is not 84 because it says an estimate for the number of lorries with a speed of more than 90. well there's a hundred lorries in total 84 up to 90 miles an hour so it's the additional 16 on top of that so it's 100 take away the 84 would be 16. but that's not just the only thing that you can find from a cumulative frequency graph and if i swap to another color we could also find the median so if i have a look if there's 100 lorries the median's going to be here sitting on the 50th and i could go along and i could find the median speed let's have a look and then drop it down it looks like that lands on 70 for me so i could have a median of 17 that's something else i could have a look at i could also have a look at the quartiles and the interquartile range again we could find 25 i could go along from there i could find 75 which is around about here i could find those two values and then work out the distance between them for the interquartile range as well so there's a lot of things that you could have a look at obviously you just need to make sure you draw it on the graph and if you do do the interquartile range which is quite a good one you go across you find your value down here again mine's not going to be perfect it looks like about 82 or 84. i should go for again the best as i can on here 84. if i go along for 25 that comes down looks like about 58 for me so i wanted to find the interquartile range i could do that as well i could do the interquartile range and that would be 84 take away 58. there you go and i'd work that out what does that come out as 26. there you go the main quartile range would be 26. my median down there was 70. but obviously this question was asking for the for the estimate there but lots of things that you can find from a humility frequency graph next up is a histogram uh this is going to be probably the fiddliest one to actually draw on here um so there's a formula that we need for his for histograms because obviously on the side of this we have frequency density so the formula for frequency density is frequency divided by class width okay well we'll have a little talk about that frequency divided by class width now essentially all we're really going to do here is find out what the height of each little rectangle is going to be for our histogram now we've already got the widths determined for us they're in the graph here so that one there 60 to 65 has a width of 5. now that frequency there is going to be the area of each bar now if you think about that if we just think logically about a rectangle if i've got a rectangle his area is 15 the width of it is five what's the height going to be the height is going to have to be 3 in order to make sure we get an area of 15 there so essentially all we're going to do is the same thing we're just going to do the frequency which is the area like in my formula here 15 and divide it by the width which is 5 15 divided by 5 gives us 3. so i'm just going to do a little separate column here for frequency density and i'm going to work out all the frequency densities so you might just want to write down these widths over here so this one's five this one's five this one's ten and then this one's twenty so we do have different widths here so 15 divided by five gives us a three which we already know the height of that rectangle is gonna have to be three the next one is 25 divided by five which is a height of five the next one is 36 divided by the 10 so a height of 3.6 and the other one's 24 divided by 20. 24 divided by 20 is 12 over 10 it's 1.2 again if you've got a calculator that's okay don't forget though some of these harder ones you could also always write that couldn't you could write 24 over 20 which is 12 over 10 and that's a little bit easier then to turn into 1.2 okay so 1.2 for that one you've then just got to be very very careful we've just got to plot this on the left so we're going to have frequency density so make sure you label that if it's not given to you our highest height there is five so i just want to get as best as i can a scale that allows me to go up to five so i'm just gonna have a quick look because sometimes these are quite hard to see so if i go up ten squares for every one just see if that fits so do this in pencil to start with just make sure your scale works we're going to start with zero and if i go up five six seven eight nine ten here let's say that's one so then up three four five six seven eight nine ten two one two three four five six seven eight nine ten three up again ten is four and then five does perfectly fit there you could do it smaller but i'm going to try and give myself as much space as possible so making sure we've got the labels here then we just actually just need to draw these in they're quite simple to draw once you've got this information so if we draw this in look just make sure you get the widths right that first speed there is 60 to 65 so my width is going to be five so it's only going to be this wide here so look is that enough one two three four another square along five and that's going to go up to three oh not very well drawn obviously using a ruler and a pencil so there's your first bar the next one has a height of five and again um and he has a width of five it goes from 65 to 70. so continuing that line up we can go up to five we'll do this as best as i can there we go there's the next one the next one has a height of 3.6 so being super accurate here and a width of 10 it goes along to 80. so if i find 3.6 as best as i can one two three four five six oh there we go and then drawing that bar in and then the final bar there has a width of 20 all the way up to 100 and a height of 1.2 so 1 is there 0.2 just about as best as i can and there is your histogram okay so we can also look at some stuff from this as well so the question here which i'm going to have a look at this part b it says work out an estimate for the number of cars with a speed more than 85 now you can actually do this from the table or you could do it from your graph now 85 is if we have a look on the graph one two three four five it's halfway between 80 and 90 there now this bar here we've already determined is 24 people or 24 lorries now if you think about this five along actually that could split up into four equal chunks that big okay it's a quarter of the way along and a quarter of 24 24 divided by four is six so in each of these little chunks we could estimate that there's six in each obviously we don't know because it is group data but we could estimate that there's six in each so if we work out an estimate for a number of cars with a speed of more than 85 well it'd be these little three chunks of six here and six plus six plus six would give us 18 and that'd be a good little estimate for us there we could also do it from the numbers so up in the table because if it's a quarter of the way along we know it's three quarters of this little group and three we could have just done three quarters of 24. three quarters of 24 is 18 divided by 4 is 6 times a by 3 is 18. so you can do little readings from your histogram like that as well so we've got our next one okay so the last one here we're going to have a look at the reverse mean question now i know it's a reverse meme when it starts to talk about actual means it says that the mean for certain people is and then starts to give us some information so it says there are 20 men 20 women and 10 children and they take part in a quiz and it gives us the mean score for the men the mean for the children and the mean for the whole group and it wants us to work out the mean score for the women so that's what we want to work out eventually so all we've got to do is think in reverse for this so when you're working out the mean you get the total and you divide by how many people there are so if we think about the men you'd have to write it like this but just so you have an understanding of this you would get the total which we don't know you divide it by the 20 bit men and it says it gives us an answer of 6 the mean for the men so if we reverse this round times both sides by 20 so times by 20 gives a score of 120 for the men so that's the total score for the men we could then do exactly the same for the children i don't have to write out this little um bit of algebra every time but if i just think okay well the children then i just know i need to times the amount of children by the mean the mean is four and there are ten children so ten times four gives us a total score of forty for the children there we go now we don't actually know what the mean is for the women that's what we're trying to work out but what we are given is this last bit of information here which says the mean for all of them is five now in total if we have a look at the question it says there are 20 men 20 women and 10 children so that's 50 people with a mean of five so that we could do 50 people times the mean which is five which gives us a total score of 250 so that's our total there so now we know the total score we know the score for the men we know the score for the children so we can actually work out what score the women got so if we do this to the side we know we have to do 250 take away 120 that the men got and take away the 40 that the children got and if we work that out 100 250 take 120 130 take away the 40 leaves us with a total score of 90 for the women so the women scored a total of 90 and it says that there are 20 women so to work out the mean for them we've got 90 is their total score divided by 20. and again depends whether you have a calculator or not here but even if you don't have a calculator we can do this one we can cancel off the zero top and bottom and it's 9 over 2 or 9 divided by 2 which is 4.5 there we go so the women have a mean of 4.5 and that'll be our final answer there to finish that off there we go and that's our little bit of statistics done okay i'm looking at some ratio in proportion now so this says james has a full bag as a bag of full bag full of counters he takes a random a sample of 20 counters marks them all and puts them back into the bag and then he takes 30 counters at random now to the 35 of them are marked estimate how many counters are in the bag so look at capture recapture here now if we just make a few statements because we're going to assume that the sample is proportional to the next one that you took as well or to the rest of the counters in the bag so initially he takes a sample of 20 counters out of we don't know that's what we're trying to work out so let's call that 20 over x on the next time he takes a sample of 30 and five of them are marked so we'll say that that's now 5 out of 30. so out of the 35 were marked but previously it was 20 that were marked out of something we don't know but we are going to assume that these are proportional so in order to work out how many that x value is we just have to think okay well how do we get from five to twenty and we do that if these fractions are equivalent by multiplying by four so in order to get that number on the bottom there we would multiply this by four and that would give us a hundred and twenty so our answer would be 120 counters okay obviously we are making some assumptions here we are assuming that no additional counters were added to the bag in between the two samples and also them were taken out as well and obviously there's little silly things like we're assuming that the marks haven't actually come off the counters or anything like that in this time frame okay so that is capture recapture there okay onto some direct and inverse proportion so there's two formulas that we can use here it says a is inversely proportional to b so it's either going to be a equals k lots of b or a equals kb or it's going to be a equals k over b so these are the two that you have to remember now with inverse proportion it's this one here where k is being divided by b so we're going to use that formula so just make sure you write them down straight away and then it gives us some values to put in so it says 15 for a so 15 equals k over and it says b is four so we can solve this for k we can times both sides by four to isolate k there and we get 60 equals okay okay equals 60. and now what we can do is we can put that 60 back into the original formula here and that will give us our formula so a equals 60 over b and that is a formula for a in terms of b and sometimes you might just be asked just to write a formula for a in terms of b and there it is but this question says find the value of a when b equals 12. so what i've got to do is stick b into the b equals 12 into my formula down here so if we do that i'm going to do it to the side we get a equals 60 over 12 and 60 divided by 12 equals 5 so our final answer there is 5. so every time you just think which formula we're going to use plug your pieces in find your value of k and then reuse that in your formula for whatever values you need okay so this question says directly proportional too so we're going to be using the other formula this time it's a equals kb but you've got to read the question carefully because this one says directly proportional to the square root of b so i need to put a square root in with my b so using the same formula but just putting my square root remember it could say squared or cubed or cube rooted or anything but that's my formula i'm going to use then it says when a is 18 b equals 16. so if we put those values in a is 18 so 18 equals k and i'm going to put a times sign in here the square root of 16 and it can do that in your head hopefully the square root of 16 is four so i'm just going to put k times four now again we can actually find the value of k here we can divide both sides by four so if i divide both sides by four we get a decimal here because it's 18 over 4 so 18 over 4 equals k now if we simplify that down let's have a look divide top and bottom by 2 we get 9 over 2 which we can either leave as that or we can turn into 4.5 i might just leave it as 9 over 2 for the moment see where we end up with this but we've got our value of k so if we put that back into our formula now so just putting that back up here for our value of k we get a equals 9 over 2. root b there we go and there is our formula for a in terms of b a is 9 over 2 or 4.5 times root b then it says find the value of b when a equals 2. so if we put those into our formula there a equals 2 so 2 equals 9 over 2 root b there we go right so we need to rearrange this so i need to divide both sides by 9 over 2 or divide both sides by 4.5 if we've got a calculator that's fine if we don't know we can just do 2 divided by 9 over 2 remember in fraction rules that'd be 2 over 1 and you would times it by the flipped version 2 over 9 and that would give us 4 over 9. so 4 over 9 equals root b and if you've got a calculator you can just type that straight in so 4 over 9 equals root b in order to find out what b is we just need to square both sides and when you square both sides there 4 squared is 16. 9 squared is 81. there we go and we get 16 over 81 so our final answer there is a fraction there we go just following those uh normal fraction rules there that would have been absolutely fine with a calculator but just a different way to approach this if it is non-calculator okay so we've got one of these questions with lots of ratios going on it says in a bag of shapes shapes with circles or squares and then they're white or black and gives us some ratios about them and it wants us to work out what fraction of all the shapes are white now if i start to break this information down a bit it says fraction so i'm going to just start to turn these into fractions now it says we've got circles and squares and that's in the ratio three to five so in total that's eight so we've got three eighths circles and 5 8 that are squares and then it gives us some information so it says the ratio of black circles to white circles is five to seven so we have a look at this we've got black and white and that is five out of twelve five plus seven and seven out of twelve for the white ones okay then it goes onto the squares so we've got one two three and again that's black to white one to three so it's one quarter and three-quarters now if i treat this just like a probability tree um because it asks me for the total fraction so if i think well what's the probability of picking a white um what is it a white a white shape so if we have a look at the circles we can go through the three-eighths and down the white shapes there to seven twelfths so just like on a probability tree we would do three-eighths and multiply it by seven twelfths and that's going to tell me the fraction of circles that are white i can do the same with the squares so 5 8 multiplied by that one there three quarters so 5 8 multiplied by three quarters and that's going to tell me the fraction of squares that are white now if we work these out times in fractions is nice and easy so times in the top let's just move this to the side a little bit so times in the top there we would get 21 over 96 times in these on the right we would get 15 over 8 times 4 which is 32. and there's our two fractions now just like with probability here we just need to add these two together which means we need a common denominator so not very nice but you might have spotted here that this fraction simplifies and if we divide the top and bottom by 3 21 divided by 3 is 7 and 96 divided by 3 is 32. so actually we can now add these two fractions together we've got 7 over 32 here and 15 over 32 here and if we just add those together to finish that off now that there's a common denominator 7 over 32 add 15 over 32 gives us a final answer of 22 over 32 and again you could even simplify your answer there if you want it doesn't say 2 but if we have both of them because they both divide by 2 we get 11 over 16. so completely up to you there we've got two different answers 22 over 32 or 11 over 16 would be a fine answer there but obviously in that scenario it's quite nice because one simplified to make a common denominator if not you just have to apply your normal fraction rules there to get your common denominator on some equivalent ratios here so when ratios are equivalent if we think about this logical idea here one to two is also equivalent to 2 over 4. i can apply a little trick here because if that is true there we can also make the statement that 1 over 2 the left over the right is equal to 2 over 4 and we know that that's correct isn't it one half is equal to two quarters i'm going to apply that same trick to this question here even though we're involving some algebra now we've got 3x plus 5 that's not a 3. we've got 3x plus 5 in the first one so 3x plus 5 over the second one there x plus 4 is equal to the first one 2x plus 4 over x plus 2. and there we go now we've actually got an equation that we can solve so this is more of an algebra question but it's a trick with the algebra there sorry with the ratio so what we can do we've got two fractions either side we can cross multiply so i'm going to times the x plus 4 up to this side and times the x plus 2 up to the top of this one so we get a double bracket we get 3 lots 3x plus 5 multiplied by that right denominator x plus two and that is equal to two x plus four what's on the top of there multiplied by that left denominator which is x plus four so all we need to do now is expand both these brackets out i'm gonna do this in quite a quick step so the first one we get 3x squared plus 11x plus 10 and on the right one here we get 2x squared plus 12x 16. so we've got quadratic c and we just need to make it equal zero i'm going to do this in quite a quick step i'm going to move everything here on the left hand side to the other side so i'm going to minus 2x squared minus 12x and minus 16 and that's now going to equal 0. so 3x squared minus 2x squared leaves us with x squared 11x take away 12x leaves us with 1x so minus 1x and 10 take away 16 is minus 6 and that equals 0 and then we can solve it so all i did was minus 2x squared minus 12x and minus 16 from both sides so that equals zero now we can solve it so looking at this now i just need to factorize it so the factors of one and six are either one and six or two and three and it's going to be three in the middle there so it'll be x plus two and x minus three and that would give us minus one in the middle and that's what equals zero so we've got two solutions here which is why it says find the possible values of x we've got this solution here you've got x could equal minus 2 and you got the one to the right of that as well you've got x could equal 3. so there's your two solutions x equals 2 and x equals 3. finish that off last question here we've got a speed time graph very similar to a velocity time graph as well similar process in what we do so part a says find an estimate for the distance traveled in the first 10 seconds we know that our formula speed is distance over time so actually working out distance is speed times time which just means all we have to do is work out the area underneath the graph so normally it would say to split this into equal widths okay i haven't actually added it to this one but we're going to split it into five equal widths so along the bottom along the time it goes up to 10 so each bar i'm going to draw in i'm going to split it at two so every two seconds i'm going to split the shape up and draw it in here we go i'm going to connect those up together here to make the shape so i've got a triangle there a trapezium another trapezium another trapezium and a final trapezium there this is going to be quite awkward for me to do on here but i'm going to give it my best shot so the height of this triangle i'm going to write them all in because the base of all these shapes is 2 but i just need to work out the area of all these shapes so the height of the triangle there as best as i can see looks like 27 like that is 27 the height of the next trapezium specifically is 44 which needs to be super accurate with these the height of the next one is 50. the height of this next one here is what looks like 44 again to me and this final height at the end here is 20. so what i need to do is work out the area of them all so i'm going to label these up one two three four and five and work out the area of all these shapes so shape number one's a triangle so base times height two times 27 and half it is 27 so just having a look at the units here is meters per second so the shape one is representing 27 meters shape two we've got the air for trapezium here don't forget the area of the trapezium formula which is add together the parallel sides or a plus b divide it by two and then times it by the height which in this case is the two d the distance between them now it's really nice when you've got a distance of two between them because actually dividing by two and terms by two cancel each other out so actually you just have to add up those two parallel sides so for number two 27 at 44 is going to give us 60 71 71 meters again plug it into this formula if it doesn't have a width of two number three is going to give us 94 44 plus 50. so 94 meters number four 50 plus all right yeah summer falls the same again 94 meters again and then number five is 44 plus 20 so 64 meters so we need to do is add them all up which i'm going to do on the calculator so adding all of those up we've got 27 plus 71 plus 94 plus 94 plus 64. and that gives us a total distance there of 350 meters all right there we go just adding them all up let's put that down the bottom there 350. okay so one more thing that we can have a look at on here and it's part b it says work out the an estimate for the acceleration at two seconds now this is a bit awkward to do on the same graph but what i'm going to do is i'm just going to get rid of some of the drawings here around the two second mark now in order to get an estimation for the acceleration we've got to draw a tangent at this point here and that can be awkward but obviously just gone up from two there to find out where that is so you just gotta get your ruler really carefully and make sure that it just touches at that point and then skims off that's very hard for me to do on the screen here but that's it drawing a tangent there and all you need to do is draw a little triangle underneath or above and work out the rise over the run or the change in y over change in x okay so if i just draw this in and a little tip of it the non-calculator is to try and get that change in x to be one so i'm just going to have a look to see if i can actually do that so if i go from here i go across one two three four five squares that's a change in x of 1 and i get that distance there which for me actually perfectly goes to 10 so it's going to be different on the paper because it's quite hard to do it on here but that gives me a 1 and a 10 and obviously to work out the gradient you do the change in y over the change in x and mine comes out as change in y is 10 and the change in x is one and it's sloping upwards so it is positive and that gives me a gradient of 10. so in terms of an acceleration here my acceleration would be 10 meters per second squared and actually from units for acceleration there so 10 meters per second squared is my acceleration right and that's the end of ratio in proportion okay so moving on to probability so we've got here emma has a bag of shapes and in the bag there are eight red seven oh sorry eight red cubes seven red spheres five blue cubes and twelve blue spheres i'm just going to take two shapes out of the bag and she's only allowed one sphere one cube one red shape and one blue shape how many different ways can she select them so look at the product rule for counting here in combinations so we have to do is think about here how many options does she have on the first pick and we're going to multiply it by the amount of options on the second pick problem is we've got this scenario here where she's only allowed one of each shape and one of each color so if we think about this at first if she takes a red cube then she can't take another red shape so she can't take any of those red spheres and she also can't take any of the blue cubes because she's only allowed one red shape and one cube so if she takes a red cube the only other option on the next pick is to take a blue sphere so for that first option there she has eight on the first pick out of the red cubes multiplied by 12 of the blue spheres so 8 times 12 would give us the amount of combinations there that she can get for the first pick and that's going to be 96 so she could take that combination there a red cube and a blue sphere now alternatively she could also take on the first pick she could take a red sphere and that would mean on the next pick she can only take a blue cube to enable her to only have one of each shape and one of each color so that would give her a total amount of combinations there of seven on the first pick there if we took a red sphere multiplied by five of the blue cubes and that'd be a total of thirty-five so she has two different ways of picking these shapes we've got 96 options on the first and 35 on the second so if we add all these different options together 96 plus 35 would give us a total amount of combinations here of 131 to meet that criteria of only allowing having one shape and one of the colors for each one there so one sphere one cube one red shape and one blue shape and that's combinations there you just gotta remember how many choices are there on the first pick can you multiply it by however many choices there are on the second pick and then combine any if there are multiple different ways okay so the next one we've got a lot of information here so it says sophia asked 50 people which drinks they like best from tea coffee and milk and all 50 people that she asked liked at least one of the drinks then it starts to talk about all these different ideas about what drinks people liked and whether they liked all three whether like tea and coffee so we're going to go through all of them there but ultimately it says she selects at random one of the 50 people work out the probability that this person likes tea so we need to figure out from all this information how many of the people there like tea and it doesn't explicitly tell us how many like tea so when we've got these three options like this the best way for us to organize this is in a venn diagram and specifically a three-way venn diagram so we need to draw i'll do this as best i can a venn diagram that's going to allow us to have three elements in there and just need to make sure the venn diagram you draw a little box around the outside there we go now i want to keep this all nice and neat and tidy so i'm going to label that i'm going to call this one the milk this one the tea and this one down here the coffee i'm just going to go through the list and just start to tick things in when i know that i've got them there now the first piece of information it tells us and if we go through this step by step it says all 50 people like at least one drink so that means we know that there's going to be nobody in the outside so zero on the outside next thing we've got is 19 people like all three drinks and that's a key piece of information there because we can definitely put that in straight away the people that like the all three are right smack bang in the middle there so we've got 19 in the middle so we've dealt with that one you can tick them off as you go or highlight them so that next one says it says 16 people i'm going to swap to a different color here it says 16 people like tea and coffee but don't like milk so the people that like tea and coffee are going to be in the crossover between the t and the c circles but they do not like milk so that's fine we can put them in there 16 that like tea and coffee but do not like milk there we go tea and coffee but don't like milk it then says 21 people that's trustworthy 21 people like coffee and milk oh i'm just going to get rid of that because i've got too many things going in there i've got 21 people like coffee and milk so 21 people that like coffee and milk well let's have a look coffee and milk is is this crossover here but also includes that 19. it doesn't say 21 people like coffee and milk but don't like tea so if 21 people like coffee and milk we've already got 19 in there that like coffee and milk so that's going to be an extra two that makes that up to 21 there in that little crossover with coffee and milk 19 of them also like tea but then 21 in total like coffee and milk so that's that one dealt with how many have we done now i've done all four of those there's lots on here on to the next one we've got 24 people like tea and milk so 24 people like tea and milk let's have a look what tea and milk is this one here but again it doesn't say that they don't like coffee so 24 people like tea and milk 19 are already in there so that's gonna be an additional five people there and that makes up 19 and five that's 24 people in total that like tea and milk on to the next one we've got 40 people that like coffee we've already got three people in our coffee circle there we've got the two we've got the 19 and we've got the 16. so in total that adds up to let's have a look 2 plus 19 is 21 at the 16 is 37 and it says 40 people like coffee so it's going to be an extra three that make that coffee circle up to 40. next one this one's nice easy one to finish we've got one person likes only milk so in the milk we've got one person and then we're given no more information but there is this missing number here in the t now just remember it did say in the question that she has 50 people so if we add up all the numbers we've got we've got 1 plus 5 plus 2 plus 19 plus 16 plus 3 and all of that adds up to make 46. and in total we have to have 50 people in our venn diagram so just in the t1 there that's an additional four people to make sure that we have 50 so we just did 50 take away 46 which left us with those remaining four people so we filled in our venn diagram now it says severe selects at random one of the 50 people work out the probability this person likes t well that doesn't say just likes t it says the probability that they're like t so we've got a few of the numbers here that we can actually consider we've got the 4 the 5 the 19 and the 16. so the props for the amount of people that like t if we add them all together 5 plus 4 is 9 plus the 19 is 28 plus the 16. what does that give us 19 plus 16 plus 5 plus 4 gives us a total of 44 people that like t so there's 44 people that like t out of a total of 50 people so it's 44 out of 50. 44 of those 50 people like tea and the rest don't like tea but the one person that only likes milk you've got the two people that like milk and coffee and the three people that only like coffee so 44 over 15 you don't need to simplify that and that's our venn diagram done just make sure you go through and tick them all off as you go so you haven't missed any let's look at another one of these okay so this venn diagram is a little bit more complicated it says 82 students whereas what their favorite fruits were 39 like apples 50 like bananas 39 for oranges 21 like apple and bananas 18 like bananas and oranges and we've got some more information there but the one we're looking for is the amount of people that like all of them and this doesn't actually tell us the amount of people like that like all of them so we're going to have to take a little bit of a different approach here but just say how many of the students like apples and oranges but not bananas so we're going to look for those people in a sec now if i draw a little venn diagram for this one again i'm going gonna need a little bit more space i might actually make that a little bit bigger let's make this one a bit bigger okay so we've got our venn diagram again nice little boxer on the outside and let's have a look at what needs to go in here so 82 students were asked what their favorite fruits were so we know that 82 people are gonna have to go into our venn diagram now it says um that we've got lots of different bits of information haven't we here so let's have just label it up so we keep it nice and clear apples bananas and oranges down here not to mistake that for zero people there that's oranges now we don't know how many people are in the middle so we're gonna apply a little bit of algebra here i'm gonna say okay well there is an amount of people in the middle and we'll call that x now we can start to have a look at some of the other bits of information i'm going to disregard the apples bananas and oranges a bit to start with but what i am going to do is just label that up here just so i remember so apples is 39 so no 39 in total in that circle bananas is 50 so in this total of 50 going into that circle and oranges is 39 so we know there's a total of 39 going into the oranges there okay so i'm going to take those all off i've drawn them on the venn diagram i'll remember that later now it says there we go 21 light apples and bananas so 21 like apples and bananas so that is going to be 21 in total including that crossover that's going in these two just here where the apples and bananas cross over so if i knew what the middle number was there i'd do 21 i'd take away the value of x in the middle but i can't i don't actually know what x is so i'm just going to write in here that's 21 minus x for that one okay so that's that one done for the moment in terms of algebra then we've got 18 that like um bananas and oranges so bananas and oranges is just here but again we don't know what that middle one is it doesn't say like like bananas oranges but not apples so that's going to be 18 take away the middle number which again is x 18 minus x onto that last one we've done this this one now so 19 light apples and oranges and again that's this one here apples and oranges but again we don't know the middle number so we'll call that 19 minus x there we go so there's our bits of algebra let's take that off and then it says 22 like exactly two of the fruits now the people that liked exactly two of the fruits are the first one we did 21 minus x i like apples and bananas we've got 18 minus x i like bananas and oranges and we've got 19 minus x like apples and oranges so what we can actually do is we can create a bit of an equation here now if we add those all together that has to equal 22 because it says 22 here like exactly two of the fruits so if we add them all up let's have a look let's just add them all up we've got 21 minus x we've got 19 minus x and we've got 18 minus x and if we add them all together let's see what we get so 18 plus 19 plus 21 gives us a total of 58 so we have 58 and minus three x's so minus three x so actually we can create a bit of an equation here i'm running out of space a little bit so i'm gonna do it over here but we have this equation we've got 58 minus three x and it says it has to equal 22 people so actually we can just go about solving this now so actually what i can do and we can rearrange it if we like but i'm just going to minus 58 from both sides so 22 minus 58 gives us negative 36 so we have negative 3x equals negative 36 and if we do negative 36 and divide it by negative 3 that gives us a value of x which is 12. so we know that x that one in the middle now has to be 12. so actually what we can do we can start to get rid of some of these numbers and actually sorry to get rid of some of these x's and replace them with numbers so if i get rid of the x in the middle and we'll replace that with the 12 because we now know that 12 is in the middle we can then do 21 minus x above that which leaves us with nine so we can get rid of 21 minus x oh do you got rid of the whole circle there we can get rid of 21 minus x and we can just put 21 minus 12 which is 9 so there's 9 going in there we can get rid of 19 minus x 19 minus 12 leave us with seven so we can get rid of that and we just got seven in there and then we've got 18 minus x on the right down 18 minus 12 leaves us with six so we can have six in there there we go six now we might be able to answer the question from here but let's go ahead and fill in the rest of the venn diagram as well so we know that in a there's 39 in total and we've got 9 twelve and seven that's already in there and nine plus twelve plus seven is twenty eight so we need an extra eleven in the apples there to make it up to thirty nine moving on to the bananas again we've got nine twelve and six and nine plus twelve plus six is twenty we want to get to 50 so that's an extra 23 and then in the bottom one we've got seven plus 12 plus six adding up to 25 we need 39 in there down the bottom so that's going to be an additional 14 right there we go so that's our venn diagram filled in now the question said not to finish what it was actually not to forget what it was actually asking us here it said how many of the students like apples and oranges but not bananas now the ones that like apples and oranges but not bananas are here let's highlight it these ones here like apples and oranges but they don't like bananas so we've got seven people there so we've got a total of seven people out of the 82 people in total and that will be our final answer there we go so you can apply this to a venn diagram that's not a triple venn diagram where you've just got a normal two two circle venn diagram as well but when it doesn't give you that crossover number there you do have to think about how you can apply algebra to this so that's quite a complicated one there but just a little example of how you can approach some of these harder venn diagram questions okay on some probability here it says there are five red pens three blue pens and two green pens in a box leah takes a random append from the box and gives the pen to a friend leah then takes that random another pin from the box work out the property that both pens are the same color so we can do this using a probability tree but there are three things so i'm going to have to have three branches here and it can get quite messy so we've got red we've got blue and we've got green now in total there we've got five red pens three blue pens and two green pens and in total that adds up to ten pens so this is going to be all out of ten pens here so the probability of getting a red is five out of ten the probability of getting a blue is three out of ten and the probability of getting a green is two out of ten now it's up to you we can do all three branches coming off these if you want for that second picket so she doesn't replace it so off each branch you've got red blue green red blue green and red blue green for the bottom but really we're only really concerned here about the same color so i want to know what's the property of red on the second pick if we take a red what's the probability of blue here if we take a blue and what's the probability of green down here if we take a green now if we think about this there are five red pens in the box so if we were to take another red pen that would reduce down to four so there would only be four red pens left and now there'd only be nine in the box so i don't need to fill in every branch on the tree here we could do we've still got three out of nine for blue and we'd still have two out of nine there for green but i only really care about the probability of getting another red at this point moving on to the blue branch so down here we have three blue pens in the box so if we take one out we've now only got two blue pens and again that's going to be out of nine so that's the probability of getting a blue if we took a blue on the first pick and then the final scenario down here if we took a green on the first pick we'd only have one green left and again there'd only be nine in the box so that'd be one out of nine so there's the probability of getting this second color when it's the same so just like normal probability trees we just need to work out these three probabilities here so for the first one we'd do five over ten multiplied by four over nine which is nice and easy for us to do times the top is twenty times the bottom's ninety into the middle we've got 3 over 10 multiplied by 2 over 9 and again that's going to be over 90 but with 3 times 2 on the top which is 6 and 90 on the bottom again and then the final one we've got 2 over 10 multiplied by 1 over 9 and 1 times 2 is 2 and again 90 on the bottom so there's our three probabilities we've got the probability of getting red red of 20 over 90 blue blue would be 6 over 90 and green green would be 2 over 90. so when we want to combine all of these together all we have to do is add them all up so add all of these together to get our total probability let's just put some plus signs here and 20 on the top plus 6 is 26 plus 2 is 28 so had final probability there would be 28 out of 90. and again you don't need to simplify that so that'd be your final probability there you go a little bit of a trick there when you are constructing these trees if you only need certain branches then you can just kind of ignore the others but just be very careful what the questions asking for because it could have asked us for different colors there just on a side note if it did ask us for different colors that would have obviously been the other six branches but you can also save your time just working out the three that we don't need and then we could take that numerator there 28 away from 90. so the probability of different colors would be 62 over 90 what's left over there different colors okay that would be only though if it said different colors and this particular question did say the same color so just be careful but you could save yourself a bit of time rather than working out all six you could just work out these three and take it away from 90 for your numerator on to the final question here on probability trees it says there are four red counters and x blue counters in a bag two counters are removed from the bag and the probability that both the counters are taken as blue is one third work out the value of x now with the tree here isn't so nasty to construct but we have got some algebra going on in the tree now what it says is there are four red and x blue now so for the total amount of counters here we don't actually know what the total is but we could write it in terms of algebra if we knew the amount of blues we just add it to the amount of reds so we could say it's x plus four is our total and that's going to be our denominator on our tree here so if we just construct two branches for our tree we've got red and we've got blue and the probability for red there is going to be four out of x plus four and the probability for blue is going to be x over x plus 4. now again really the only problem part of the tree here we're actually considered you know we want to consider is the probability for both of them being blue because it says the probability that the both counters of blue is a third so i don't really need to do the whole tree here i just need to have a look at this probability down here and that would be if i took a blue on the second pick but again i'm actually just going to go and construct all the branches here so that we can just discuss them but we only actually need the one at the bottom now if we took a red that would reduce our amount of reds down to three because there are four reds in there at the moment so if i imagine that we've taken a red out we would only have three and that would no longer be over x plus four because one of them's gone so it'll be over x plus three reducing that denominator by one again onto the blue branch if we take um one of the reds out i've seen one none of the blues have actually gone so we've still got x blues but that is going to be over x plus three again all the denominators are the same onto the bottom half of the tree if we took a blue out on this bottom half the amount of reds would be the same so we would have four still over x plus three because we've lost one on the bottom and the one that we're really concerned about here is this bottom one which is if we took a blue out we don't know how many blues there are so we'd have to just say okay well that'd be x minus one counter over x plus four and that's how we can construct a probability tree using using algebra here now what it says is it says that those probabilities down here so the probability of getting a blue and then getting another blue is one-third so when you times these fractions together it says they're going to equal a third so if i try and get rid of a lot of this tree here because we're only really considered concerned about these bottom ones let's have a look at what we're left with so these are our branches so we can actually do the working out for this now what we're saying is x over x plus 4 the first probability multiplied by x minus one over x plus i just realized plus four done that should be a plus three oh dear let's change that x plus three there we go because we took one of the counters out so x minus one over x plus three and it says that all of that equals a third and obviously i've got limited space here so i'm gonna get rid of that tree now that we've corrected that denominator and we've just gotta actually work out all of this algebra here we've got a bit of an equation so if we go about working this out now if we multiply these fractions together multiplying the numerators x times x minus 1 gives us x squared minus x and if we multiply the bottoms there we have a double bracket to work out okay so you can imagine these are in a double bracket that first one was like a little single bracket x times x minus one so if we multiply out the double bracket here we get x squared plus four x plus three x which is plus seven x and four times three is twelve so plus twelve and it says that that equals one third okay so from here we've got to cross multiply uh to get rid of these two denominators so i'm going to times the denominator over here up to the one and the three up to the other side there so times both sides by 3 and times both sides by this quadratic on the bottom the x squared plus 7x plus 12. so if we do that i'll write this out as a like a step you can see to start with we'd have 3 lots of x squared minus x equals one lot off so i won't write this in a bracket but one lot of x squared plus seven x plus twelve and now we can actually go about solving this so obviously we need to expand this bracket out to start with so both times these both by three and we get three x squared minus three x equals x squared plus seven x plus twelve now just like with all quadratics we need to make this equal zero so i'm gonna move everything to the left hand side so i'm going to subtract all three of these pieces i'm going to subtract x squared which would leave us with 2x squared when we take x squared from both sides i'm going to take away 7x from both sides which would leave us with negative 10x over here i'm going to take away 12 from both sides we haven't got a number on the left such it's going to be take away 12 there and now it equals zero now we can actually go about uh solving this so we can either factorize or use the quadratic formula this one i think does actually factorize but we can simplify it first we can divide everything by two so if we divide everything by two we get x squared minus five x minus six which equals zero and now it's quite nice and easy for us to factorize there so i'm gonna bring this up to the top i'm going to factorize it out of the way up here so factors of six are one and six or two and three so this actually factorizes let's do it up here open up our brackets and we want minus 5 in the middle plus a negative 6 at the end so the only option in there is going to be minus 6 and plus 1. that would allow us to get minus 5 so we'd have x minus 6 and x plus 1. there we go we factorized it so we've got our two solutions there so obviously you flip the sign so x is six and for here x equals negative one so what does this mean in terms of the question now it said work out the value of x this is what we've been doing here working out the value of x and x there is the amount of blue counters now there can't be minus one blue counters can there but there can be six blue counters so it has to be the positive value there x would have to equal six as our final answer because we can't have minus one blue counters so just something to be thinking about there right there we go and that's the end of probability right well done for making it to the end of the video here are all the topics that you've covered coming up make sure that you do some further revision on some of these topics if any of them with ones that you've highlighted is things that you need to have a further look at but while i don't get into the end there's a lot of content there hopefully it's a really good help for you pre-exams to cover off all these topics if you are aiming for those higher grades but again do do your own subject analysis make sure that you have a look through your revision guide just don't forget if this was usual useful please do like the video please share it with your friends and obviously subscribe to the 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