okay here's a video that's going to review all main concepts you learned in grade 9 math in under 60 minutes you're going to learn all grade 9 in under an hour this is a great video if you're doing exam review or if you're about to take the course and you just want to know what you're going to learn now in this video I'm not going to go in-depth into any of the topics I'm going to go through a couple examples of each topic but if you want in-depth explanations and proofs of each topic go to Jensen math GA and there's video lessons for each topic there so I'm going to divide this video into three parts based on the three main units you do in grade 9 so the first part is going to be on algebra and algebra you learn about polynomials you learn how to collect like terms you learn distributive property learn exponent laws and the next part is solving equations so I'm going to go through that part first and watch the second video if you want to learn about linear relations and a third video if you want to review the geometry section so watch all three videos and you'll have reviewed all of the main concepts of grade 9 math in under an hour so let's get started with the first topic right away I'm going to go quickly just so I can get this done under an hour so the first main topic you would have learned in grade 9 math is exponent laws so to do exponent laws first main thing you have to remember you have to know what a power is so let's say we have five squared we have to know that that means five times five right this is the base five is the base 2 is the exponent and this is a power so the exponent tells you how many of the base so how many factors of the base are being multiplied together so there are some exponent laws you would have learned the product of powers rule so if you have two powers being multiplied together the dot means multiply so two powers being multiplied together have the same base that's important the base has to be the same there's a rule you keep the base the same and you add the exponents so that equals x to the a plus B so you can rewrite as a single power by keeping the base the same adding the exponents what if you have two powers the same base being divided by each other there's a similar rule for that one but similar rule you keep the base the same and this time you subtract the exponents that's X to the a minus B you would have learned the power of the power rule so if you have it exploring on top of an exponent can keep the base and you multiply the exponents X to the a times B what if you have a power where the base is the quotient so a power of a quotient you have to remember this exponent gets put on to the numerator and the denominator so this is equal to a to the x over B to the power of X what if we have a power where the base is a product of numbers or variables so this exponent outside of the brackets gets put on to each of the factors inside the brackets so this would be equal to a to the x times B to the X and a couple more rules you would have learned that anything to the power of zero is equal to one and the negative exponent rule if you have power with a negative exponent you can't leave a final answer with a negative exponent in this section so what you have to do is rewrite this the positive exponent by writing the reciprocal of it so it would be 1 over X to the power of positive a so I'm going to go through a couple examples that review all these rules that these are the main exponent rules you would have learned in grade 9 so let's practice a few of them without variables first and I'll throw some variables into the mix so here we have powers that all have the same base right y'all the base of for the first two are being multiplied together so what I can do is rewrite as a single power by adding the exponents 3 plus 5 is 8 and then I have to divide that by 4 squared so I can rewrite that since they have the same base and they're being divided by keep the base the same and I subtract the exponents so I end up at 4 to the power of 6 and then after I've written it as a single power I can evaluate it for the power of 6 is 4096 number for the power of 6 means 4 times 4 times 4 times 4 times 4 times 4 it doesn't mean 4 times 6 okay power of a quotient and we have power of a power here as well so let's start off with a power to power rule so I have an exponent on top of an exponent I can rewrite this by multiplying these powers together 2 times 3 is 6 and notice I put the negative sign into the numerator I could have put it into the denominator if I wanted to you but don't put it into both I decided to put it to the numerator to make it easier for me here so here I have negative 2 over 3 to the power of 6 so what I have now is power of a quotient so remember 4 power of a quotient the exponent if the base is a quotient the exponent of the power goes onto the numerator and the denominator so this equals negative 2 to the 6 over 3 to the 6 and then I have to let me move this down and then I have to evaluate each of those powers so negative two to the power of six is 64 and three to the power of 6 is 729 so that's fully simplified I can't reduce that fraction any further let's look at these two here so I have product of powers here and the powers have the same base so I can use my exponent want to simplify by keeping the same base and adding the exponents 4 plus negative 7 be careful with your integers 4 plus negative 7 is negative 3 but I can't leave that as my final answer with the negative exponent what I have to do is rewrite that with a positive exponent by writing the reciprocal of it so 1 over 5 to the power of positive 3 and now I can evaluate 5 to the power of 3 and that gives me 125 so 1 over 125 is my final answer there how about here I have quotient of powers so what I have to do I can rewrite it as a single power by keeping the base and subtracting the exponents 7 minus 7 0 and remember anything to the exponent of 0 is equal to 1 and you should notice up here when you have something divided by itself that answer is always going to be one as well so think of it either way let's throw some variables into the mix so I have a product of powers here I have these three powers being multiplied together they're all the same base so I can rewrite as a single power by keeping the same base and adding the exponents 3 plus 4 plus 5 that's 12a to the 12 I can't evaluate it evaluate that since it is a variable sweet leave it like that there's our final answer what we have here is power of a product so the base of the power is a product of 4x squared and Y to the 5 so what I have to do is put this outside exponent onto each of the factors in the product onto the 4 onto the x squared and on Y to the fly so I have to do 4 cubed I'll put in brackets just so it looks nicer here I have to do x squared cubed and I have to do Y to the 5 cubed write this outside exponent goes on to each of the factors of the product of the base and now I can evaluate 4 cubed is 64 X to the 2 to the 3 well that's power the power I don't have to multiply the exponents there so that's X to the six and Y to the five to the three power of a power rule again Y to the 15 and that's done here I have product of some powers and of some coefficients here so let's start off with the five times four always start with your coefficients we can multiply five times four just as you always would it's 20 now you look for powers that have the same base I have an M to the five and an M Squared those have the same base so I know when I multiply powers to add the same base I keep the base the same and I add the exponents five plus two is seven so I've m2 the seven also I have two powers of M that are being multiplied so I can simplify that by writing as a single power by adding the exponents and keep in mind this M has an exponent of one even though you don't see an exponent there so one plus four is five and that is fully simplified there that question is done let's look at a quotient of powers here same thing stare with the coefficients start with your 36 divided by 27 and divide 36 by 27 exactly as you always would we don't want a decimal answer we just want to reduce it right 27 doesn't go into 36 evenly but I can reduce that fraction by finding number that goes into both of them evenly and in fact nine goes into both 36 and 27 9 goes into 36 four times and it goes into 27 three times so I can reduce 36 over 27 to four over three now I look at my variables so I have some powers but with the same bases I have an X cubed and X to the six that are being divided by each other so I know I can simplify that writing as a single power if I subtract the exponents 3 minus 6 is negative 3 and make sure you put your quotients into the numerator here we'll take care of that negative in a second next we also have two powers of a base of why Y to the 9 divided by y to the 4 rewrite it's a single power by keeping the base and subtracting the exponents 9 minus 4 is 5 and remember always be your quotients in to the numerator now we have to do something with this answer we can't leave our answer with a power that has a negative exponent this X to the negative 3 is a power with a negative exponent we can take care of that negative by writing the reciprocal of it and what's going to happen is this X to the negative 3 not before just the X to the negative 3 just that power if we bring it into the denominator it'll make the exponent positive so what we have for our final answer the 4 and the y to the 5 stay in the numerator the 3 stays in the denominator and we bring the power of X to negative 3 to the denominator and it makes it a power of positive 3 so there's my final answer there let's do one more if you can do this example you can be confident that you understand all of the exponent laws so for this example start with just the numerator forget about the time therefore now let's look at just the numerator and simplify that I have an 8 times a 4 that's 32 I have now look for powers the same base of a b cubed times a B to the 1 keep the base add the exponents remember product rule when multiplying powers to in base you add the exponents and now I have a d to the 1 times a D to the two that's D to the 3 if we look in the denominator let's just leave this two out front for now and let's just do this power of a product rule here where this exponent of 2 has to go on each factor of the base has two on the to the B and the D so we have to square the two which makes it a 4 we have to square the B which makes it a B squared and square the D which makes it a d squared so now I have 2 times 4 B squared d squared so I can simplify that 2 times 4 and make it an 8 so be the 4 leave the numerator for now 2 times 4 8 and then at b squared d squared and last step let's do or dividing 38 32 divided by 8 divide those just like you would divide any old numbers 32 divided by 8 goes into 32 4 times so my answer for now do B to the 4 divided by B to the 2 when dividing powers the same days keep the base subtract the exponents D to the 3 divided by D to the 2 once again keep the base subtract the exponents it's a d to the 1 if it's a 1 you don't have to write the 1 alright that's it for exponent law review let's move on to polynomials quickly so basically you have to remember what a term is a term is an expression form of a product of numbers and variables so for example 2x that's a term it's a product of a number 2 and a variable X and what a polynomial is such an expression interesting on one or more of these terms that are connected by addition or subtraction operators so for example I could have the polynomial 4x squared plus 3x plus 1 that's a polynomial where we have three terms term 1 2 3 they're separated by addition or subtraction signs now we can classify polynomial based on how many terms as by name if the polynomial only has one term like this 2x up here we call that a monomial if a polynomial has two terms we call it a binomial if it has three terms we call it a trinomial and if it has anything more than that there's no special name for it like that's four terms we just simply call it a four term polynomial if it had five we would call it a five term polynomial and so on so one more thing you'll have to be able to do is state the degree of a term and the degree of a polynomial now just state the degree of a term by looking at one individual term you can state the degree of it just by adding up so finding the sum of the exponents on all the variables in that term so if we look at this first example here three x squared Y this is one term so this is a monomial there's nothing else added or subtracted from it it's one term has two variables so to find the degree of this term we add the exponents on the variables Q minus y has a one so on the X and on the Y we add the exponents 2 plus 1 is 3 so this this right here this term right here is degree 3 if we look at this next example here what we have here is three different terms being added together so we call this a trinomial we could find the degree of each term like this first term here is degree 5 this first term here is degree 4 because that has an explorative one and this third term is degree 6 so if we want to figure out the degree of the entire polynomial all we have to do is figure out the degree of the highest degree term in this polynomial we don't add all these degrees together we just pick the term that has the highest degree so in this case it would be the third term its degree 6 so what we say is the entire polynomial is degree 6 the degree of the polynomial is equal to degree of the highest degree term in the polynomial here we have a binomial right two terms one to subtract separated by a subtraction sign the first term is degree seven right one plus five plus one is seven this term is degree six so to find the degree of the entire polynomial it's equal to the degree of the highest degree term this term is the higher degree so the degree of the entire polynomial is seven let's look at collecting like terms now so first of all what are like terms like terms are terms of the exact same variables with the exact same exponents for example these two terms they have the exact same variables they both have an X and a Y and on those variables are the exact same exponents on the X's of 2 and on the Y is a 1 on both of them so those are like terms the fact that the coefficients are different do not make them not like terms the coefficients don't matter when deciding if they're like terms or not you just look at the variables and the exponents so if we look at these two here these are not like terms why because this one has a Y to the power of one this one has a Y to the power of two so they don't have the same variable so the same exponents so they're not like terms why do we have to be able to know what like terms are because we can we can collect like terms together for example here I have four terms so when better being added or subtract from each other when adding or subtracting terms we can collect them together by adding or subtracting the coefficients only and keeping the variable the same right this is different than exponent laws with exponent laws we were using them when we are multiplying powers or dividing powers but now we're going to be adding or subtracting terms and we can collect them we can collect like terms together so what we want to first do is group the terms they're like terms together and make sure the sign that is to the left of the term stays with it so for example I have a negative 2x and a negative 5x those are like terms so let's write those beside each other keeping the signs that are to the left of them and I also have a positive seven line a negative 9y let's write those beside each other because those are like terms now I can collect them together I can collect the negative 2x minus 5x together because they're like terms by just adding tracking the coefficients only so negative 2 minus 5 thats negative 7 and then you keep the variable exactly the same don't change the exponent on the variable at all it stays exactly the same don't get this confused with exponent laws now look at this group of like terms the 7y minus 9y 7 minus 9 is negative 2 so I write negative 2y so that is fully complete I can't collect these two terms together because they're not like terms so that expression is fully simplified let's look at the next expression here it's a little bit longer find the groups of like terms I have a 3 x squared Y and an 8 x squared Y those both have the exact same variables with the same exponent so I'm going to write those beside each other 3x squared y 8 x squared Y I have a 4y and a negative 1y so I'll write my positive 4y a my negative 1y remember the coefficient is 1 if you don't see it and I also have two constant terms constant term means the term without a variable those are like terms I have a positive 7 and a positive 80 right those beside each other now collect your like terms so 3x squared y plus 8x squared Y just add the coefficients only 11x squared Y don't change the exponents on the variables at all when you are adding or subtracting like terms together I have a positive 4y minus 1y that's positive 3y and I have positive 7 plus 80 that's positive 87 that expression is fully simplified none of these three things are like terms with each other so I can't collect them together don't try and go any further than you can and notice the order I wrote these in you should do it in this order the highest degree term goes first and then it goes in descending order this term is degree 3 right 2 plus 1 is 3 this term is degree 1 and a constant term is degree 0 so the degrees should go in descending order okay it's like a distributive property so basically for multiplying a monomial by a polynomial in this case a binomial this is how you do it everything inside the brackets gets multiplied by the term out front of the brackets and that gets rid of the bracket so if I want to do a times X plus y I have to do a times X right here and I have to do a times y here so for example if I 5 times 4x plus 2 I need to multiply the 4x and the to both of them by the five that's out front so five times four X is 20 X 5 times 2 is positive 10 there's my solution those can't be collected together because they're not like terms so let's practice distributive property so here I'm doing a monomial multiplied by a trinomial so everything in the brackets needs to be multiplied by the term out front be careful with your signs and then that will get rid of the bracket so I have to do negative 3 times 2x squared well negative 3 times 2 is negative 6 so I have negative 6x squared I have to do negative 3 times negative 5 X don't forget this sign belongs to this 5x so negative times negative is positive 15 X and after your negative 3 times positive 4 that's negative 12 and I can't collect any of these 3 terms together because they're not like terms how about here I'm going to have to multiply the X plus 3 by the 3 out front I'm going to have to multiply this X and this one by the 2 up front that will get rid of the brackets so 3 times X is 3x 3 times 3 is positive 9 2 times X positive 2x 2 times 1 positive 2 now I can collect like terms here because I have a 3x and a 2x put those together that's 5x and I have a 9 and the to put those together that's positive 11 let's do one more example here for collecting like terms I have 4 K times K and times negative 3 now keep in mind this K inside the brackets think of that as a 1 K if you want so when multiplying 4 K by 1 K you multiply the 4 and the 1 together that's 4 multiply the K and the K together with multiplying powers the same base you keep the base add the exponents there's a 1 on both of them 1 plus 1 is 2 4 K times negative 3 is negative 12 K right positive times negative is negative over here I have 2 negative 2 times K squared that's negative 2 K squared I have 2 negative 2 times negative 3 K negative times negative is positive 6 K and I also have to do negative 2 times positive 4 that gives me negative 8 here if you don't see a number in front of the brackets there's an invisible one there so I have to do negative 1 times K squared negative 1 times negative 5 so that gives me negative 1 K squared and negative 1 times negative 5 is positive five collect your like terms to the highest degree terms first I have a 4 K squared negative 2 K squared negative 1 K square let's collect those together 4 minus 2 is 2 minus 1 is 1 so I have 1 K squared I'll just write that as K squared next I have a negative 12 K a positive 6 K negative 12 plus 6 is negative 6 all right negative 6 okay and lastly my constant terms I've got a negative 8 plus 5 negative 8 plus 5 is negative 3 so I'll write minus 3 and that's done I can't collect any of those three terms together because they're not like terms you probably have learned this section before distributive property we learn adding and subtracting polynomials I think it's easier to do this section after you know distributive property so basically if you have a set of brackets and you don't see a number up front there's a 1 there here there's a 1 there there's a 1 there now to get rid of the brackets you multiply everything in the brackets by the number out front so 1 times X and 1 times negative 6 that's not going to change anything that's just going to give us X minus 6 but here if there's a negative 1 out front multiplying both by negative 1 it's just going to change the sign of both terms in the brackets negative 1 times 2 is negative 2 negative 1 times negative 5x is positive 5x so what happens is both of these terms change side the signs change here I have positive 1 times X positive 1 times 4 that's not going to change anything multiplying things by 1 doesn't change anything collect your like terms I have a 1 X plus 5x plus 1 x that is 7 X and now my constant terms have negative 6 take away 2 plus 4 negative 6 take away 2 is negative 8 plus 4 is negative 4 so I have minus 4 and that expression is fully simplified can't collect those together because they're not like terms ok let's move on to the last thing you would have learned in the first unit in grade 9 you have learned how to solve equations and basically a solving equation means to figure out what value of the variable makes the equation true so here we have X plus 4 equals 7 that means something plus 4 is equal to 7 now you probably guess the answer is 3 because 3 plus 4 is 7 and that's the right answer but for more complicated questions like when we get to questions like here here you're going to want to be able to solve these algebraically using probably the balance method as a method your teacher would have taught you first so basically you're allowed to solve this equation figure out the value of X makes it true by isolating the variable by moving all of the other numbers away from the X onto the other side of the equation until you have X by itself and then you'll have your answer so to isolate the X we want to move everything away from it so right now there's a plus 4 on the same side of the equation with it we can remove that plus 4 by subtracting 4 as long as we do the same thing to both sides of the equation that keeps the equation balanced and we're allowed to do anything we want to the equation so long as we do the same thing to both sides I've subtracted 4 from both sides so the equation is still equal both sides are still equal to each other because I've done the same thing to both sides and look what happens if we subtract 4 from both sides on the left I have 4 minus 4 that's 0 so all that's left on the left side of the equation is now the X on the left side of the equation I have 7 minus 4 and 7 minus 4 is 3 so 3 is the correct answer and don't forget you can plug this answer back into your equation to check your work is 3 plus 4 equal to 7 yes you have the right answer now you should notice that a trick to figure out how to isolate the variable right now this g5 is being subtracted from it used to the opposite of subtracting 5 which is adding 5 so if we add 5 don't forget to do it to both sides of the equation whatever you do to one side you have to do to the other if I add 5 to both sides of the equation on the Left I've negative 5 plus 5 that's 0 so all that's left on the left side is G and on the right I have negative 3 plus 5 negative 3 plus 5 is 2 so that's my final answer G is equal to 2 and you can check your answer by plugging it back into your equation is 2 minus 5 equal to negative 3 yes so we have the right answer here this is different because it's not 5 plus u it's five times U we have five use so it's isolate the U that is currently being multiplied by five we do the opposite of multiplying by five which is dividing by five and remember you have to balance the equation whenever you do one side you have to do to the other and then on the left side of the equation we have five divided by five which is one so those cancel out so what you have left you have just a you on the left side and on the right we have negative 20 over five and what is negative 20 divided by five it is negative four don't forget you can check your answer 5 times negative 4 is negative 20 so it's the right answer okay let's move on to two step equations first thing you want to do is you want to isolate the term that has the variable so we want to isolate the 7y by moving this positive eight to the other side by subtracting eight so what we're going to do subtract 8 from both sides of the equation and on the Left we have 8 minus 8 that's zero it's gone so all we have left on the left side equation is 7y on the right we have 15 minus 8 and 15 minus 8 is 7 so now we have 7y equals 7 right now the y is being multiplied by 7 to separate a coefficient from a variable like this we have to divide both sides by the 7 and these 7s on the Left cancel out because 7 divided by 7 is 1 so all we have left is y equals 7 over 7 and 7 over 7 is 1 so there's our final answer you can double check if we plug 1 into this equation 8 plus 7 times 1 is 15 we have the right answer okay what if we have an equation where we have more than one term that has a variable what you want to start by getting both of those terms the same side of the equation now I want to point something out here for these previous questions what you might have noticed you could have done instead of using balance method you could have thought of just moving things to the other side of the equation by doing the opposite operation like we can move this plus 4 over by making it a minus 4 right you can move something to the other side of the equation as long as you apply the opposite operation right and that would give us what we had 7 minus 4x is 3 so let's look here we want to get all the terms with the same width with a variable in the same side of the equation so I want to move actually I'm going to move the the other way I'm going to bring the negative 8x to the right side of the equation and get all the terms that don't have a variable to the other side so I'm going to bring the plus 15 to this side by applying the opposite operation so that just means we're going to change the sign of the term so on the left side equation I'm going to leave the negative 5 and bring this +15 over and it becomes a minus 15 on the right side of the equation I'm going to leave the 2x I'm going to bring this negative 8x over which is going to change the sign of the term becomes a plus 8x now collect my like terms and then isolate the variable right now the X is being multiplied by 10 the opposite of multiplying by 10 divided by 10 so divide both sides by 10 to keep it balanced the tens cancel and what I have is negative 20 over 10 equals X negative 20 divided by 10 I know that's negative 2 so negative 2 is my answer and you could plug that back in and check your answer here if you have brackets let's start off by getting rid of the brackets by using your distributive property that you would have learned previously so start by getting rid of the brackets by multiplying the term out front by all the terms inside the brackets notice I didn't distribute to this negative because it's not in the brackets so I have 4x plus 12 equals 2x plus 12 minus 8 now we want to I'm going to simplify this 12 minus 8 if you don't mind positive 12 take away 8 that's positive 4 so I'm just going to simplify that quickly I'm going to get all the terms with the variable on the same side so I'm going to bring this positive 2x to the left becomes a minus 2x I'm going to bring the constant terms to the right so the positive 4 stays on the right bring the plus 12 over becomes a minus 12 collect and then isolate the variable the X is being multiplied by Q so divide both sides by 2 these twos cancel because 2 divided by 2 is 1 and what I have I have just an X on the left on the right I've negative 8 divided by 2 which is negative 4 and you can don't forget you could double chance check your answer by plugging into the equation here to make sure you have the right answer okay here let's look at fractions so I have C divided by 3 well don't forget if we want to move that divided by 3 to the other side what's the opposite of dividing by 3 well the altitude divided by 3 is multiplying by 3 so I'm going to multiply both sides of the equation because we have to keep it balanced by 3 so I rewrote the equation but multiply both sides by 3 why did I do that because here I have a 3 divided by a 3 most ones they cancel out so all I have left is C equals three times 2 which is 6 and we can check is 6 divided by 3 equal to 2 yes we have the right answer what if it looks like this whenever you see a fraction in an equation you can get rid of the fraction by multiplying both sides of the equation by whatever the denominator is so I'm going to multiply the left side by 4 and I'm going to multiply the right side by 4 whatever you do one side you have to do to the other and why did I choose 4 because 4 divided by 4 is 1 so they cancel out so what I have is 1 times X minus 3 which is just X minus 3 on the other side 4 times negative 2 which is negative 8 move this minus 3 to the other side or think of balance method opposite of subtracting 3 is adding 3 so add 3 to both sides on the Left negative 3 plus 3 is zero so I just have x equals negative 8 plus 3 negative 8 plus 3 is negative 5 let's do another question with multiple fractions so we have multiple fractions here if you have more than one fraction you can get rid of both of them at the same time if we multiply both equations by a common denominator so what's a common multiple between 3 & 5 so if we were to count by threes and count by 5's what's the first number they would have in common it would be 15 so what we're going to do is multiply both sides by 15 remember whatever you do to one side you have to do to the other side to keep it balanced so I multiplied both sides by 15 and you'll see how that works right here on the left I have 15 divided by 3 15 divided by 3 is 5 on the right I have 15 divided by 5 15 divided by 5 is 3 so now my fractions are gone so I have 5 times 1 times X plus 4 so I'll just write 5 times X plus 4 and here I have 3 times 1 times X plus 2 so 3 times 1 is 3 so 3 times X plus 2 and now you know how to solve it from here I'll just do it very quickly get rid of your brackets using distributive property 5 X plus 20 equals 3x plus 6 get all the terms with the variable on one on the same side I'll bring them to the left so I'll bring the 3x over becomes 2 minus 3x leave the constant of 6 bring the plus 2000 comes to minus 20 I have 2x equals negative 14 whoops and minus 22 x equals negative 14 and then isolate the X by dividing both sides by 2 because that X is being multiplied by 2 right now make sure it stays balanced those twos cancel and I have x equals negative 14 over 2 which is equal to negative 7 so there's my final answer x equals negative 7 what I have here for this next example I have a special case that we can use here now we could multiply both sides by 30 to get rid of the brackets that works fine but I have a shorter way if we have fraction equals fraction and that's it nothing else add it off at the end here anywhere else if you have fraction equals fraction you can use a shortcut you can use what's called cross multiplication you can multiply the denominator of one side by the numerator of the other so 6 times X minus 4 is equal to on the other side of the equation write the product of the denominator of the other fraction by the numerator of the other one so 5 times X minus 3 and then solve from here distributive property 6x minus 24 equals 5x minus 15 get all the terms of the variable on one side 6x minus 5x equals negative 15 plus 24 and we get x equals 9 there's our final answer you could plug back in and check now cross multiplication is nice it's a nice shortcut but make sure you only use it when you have fraction equals fraction that's the only time it works this if you can do this question you're good by solving equation so at this question here what we need to do is get rid of all three fractions and on one side of the equation we have multiple terms so we're going to have to find a common denominator between 2 3 & 4 and that would be 12 so we're going to have to multiply both sides of the equation by 12 if we want to get rid of the fractions so I'm going to multiply both sides by 12 and what happens over here since I have more than one term on this side this 12 is going to get multiplied by both terms so what's going to look like I'm going to do 12 times X plus 1 over 2 plus 12 times 2x plus 3 over 3 equals 12 times x over 4 and then from there we can simplify 12 divided by 2 is 6 12 divided by 3 is 4 and 12 divided by 4 is 3 so I have no more fractions I have 6 times X plus 1 plus 4 times 2x plus 3 equals 3 times X which I'll this ray is 3x and then you're good solving from here distributive property it X plus 12 equals 3x I'm going to collect some like terms on the left here for 6x plus 8x is 14 X 6 plus 12 is 18 I'm going to get all the terms of the X on the same sides I'm going to bring the positive 3x over becomes a negative 3x bring the plus 18 over becomes a negative 18 so I have 11x equals negative 18 and what I have here divide both sides by 11 to get the X by itself and what I have is x equals negative 18 over 11 and that's a fine answer a fraction is a perfectly fine answer just make sure it can't be reduced any further X is negative 18 over 11 okay last thing probably you would have done in the solving equations is look at some word problems I'm going to do a very quick one just to remind you how to do it so to friends or collecting Popkin tabs and Tasha has 250 more than Kristen together they have 880 Popkin tabs how many Popkin tabs to each friend has each friend collected so the question wants to know how many pop-can tabs does Natasha have how many pop-can tabs does Kristen have so that's what the question is asking us so we want to start by making a polynomial expression for each of those people Natasha is 250 more than Kristen we don't know how many Kristen has so we use a variable for Kristen and we noted Tasha has 250 more than X so X plus 250 represents that scenario and then we write an equation using these two expressions together they have 880 so I know if I add those two expressions together so if I do X plus X plus 250 I know that that should equal 880 now I can solve this equation 1 X plus 1 X is 2x bring that 250 to the other side becomes a minus 250 so what I have is 2 x equals 630 and then divide both sides by 2 to get rid of that coefficient of 2 I have x equals 315 so what does that tell us it tells us kristen has 315 and Natasha member is 250 more than X and we know X is 315 so what I have is 565 and I know together those two numbers add to a so I have the correct answer here okay that's it for algebra watch the next two parts watch part 2 linear relations in part 3 geometry geometry will be the shortest section this is the longest section of grade 9 so make sure you watch the next two sections and you'll have learned all of grade 9 in under one hour