Transcript for:
Math Final Exam Tips

let me take you through this grade 9 math final exam i've given to classes in the past it's out of 107 marks it starts off with a multiple choice section and then there are short answer sections on algebra linear relations and geometry i'll show you the full solutions to all the questions and i'll give you some tips on how to successfully write math exams [Music] question number one 87 to the power of zero equals i know anything to the power of zero except for zero to the power of zero is equal to one so c and for a question like that where it's a simple evaluating question type it on your calculator double check your answer question two a to the power of ten times a to the power of 5 written as a single power okay well i know when multiplying powers with the same base you keep the base and add the exponents so we get a to the power of 10 plus 5 which is a to the 15 and i see that as an option c again question number three i have a power of a power where the base of a power is a product of two things i know the exponent rule tells me i put the outside exponent on both factors of the base so i have to square the 3 and i have to square the x to the four three squared is nine x to the four squared and exponent on top of an exponent you keep the base and multiply the exponents four times two is eight so i'll find the answer nine x to the eighth here it is here c now for questions like two and three that involve variables you can still check your answer anytime you're saying that two expressions are equal to each other like we're saying that a to the 15 is equal to a to the 10 times a to the five if we're saying two things are equal pick a test value for a sub it in and make sure those two expressions are actually equal to each other so let's check for let's pick a value for a how about three so i'm verifying question number two right now so i wanna see is 3 to the 10 times 3 to the 5 is that equal to 3 to the 15. if we've chosen the correct answer it should be equal let me check my calculator there we go i verified it's true for three so we probably have the correct answer you can follow the same logic for verifying your answer to number three pick a value for x plug it into both expressions make sure they're the same on to number four seven to the power of five times seven to the power of well if i don't see an exponent i know it's one divided by seven to the power of three written as a single power well let me combine the first two powers and when multiplying powers with the same base you keep the base and add the exponents 5 plus 1 is 6. that product now needs to be divided by 7 to the power of 3. those powers again have the same base so keep the base and when dividing powers to the same base you subtract the exponents 6 minus 3 is 3. so seven cubed and this one will be easily verifiable by typing both expressions on your calculator and making sure they're equivalent number five i have a quotient of two expressions i want to simplify it so i'll start with the coefficients eighteen divided by six that's three and now i'm going to look for powers that have the same base that are being divided so i have two powers of m being divided so i keep the base and subtract the exponents six minus two is four and i have two powers of n being divided subtract their exponents 1 minus 1 is 0. means we have 0 n's so we could just erase that power of n or if you want to use your 0 exponent rule remember n to the power of 0 is one so three m to the four times one is just three m to the four and here we have it three m to the four t six i have a power where the base is a fraction the exponent applies to the numerator and the denominator so i have to square the 1 and the 4 that gives me 1 over 16. here's my answer here a there we go we're done the first page of the exam and almost all of these are easily verifiable with your calculator number seven this one we wouldn't be able to verify with the calculator we just have to know what like terms are question seven says which pair of terms are not like terms so like terms have the exact same variables with the exact same exponents as i go through these part a both terms have a to the 1 b both terms have an m and an n variable but the exponents on the ends are different the first one is an n to the one the second one is an n to the two so these ones are not like terms i would just quickly double check c and d notice c p squared q one p squared q one and d x to the one x to the one so we have our correct answer b b are not like terms notice i didn't care what the coefficients were for any of these we're just looking at the variable part of each term number eight this expression is a oh it wants us to classify by name so we need to know how many terms are there and the terms are separated by the addition and subtraction signs so these are the separators and i see one two three terms that means it's a trinomial nine that same polynomial expression wants to know what's the degree of that polynomial expression well to find the degree of a polynomial you need to know that it's equal to the degree of the highest degree term so we need to find the degree of all three terms the first term to find its degree we add the exponents on the variables four plus one is five the second term one plus three is four and the last term six what's the highest degree term it's this one its degree is six so the degree of the polynomial 6. and of course your teacher is going to put common mistakes as possible answers where you have to remember that you don't add the degrees of all the terms right it's not 15 it's equal to the degree of the highest degree term 6. 5 to the power of negative 2 is equal to well i know the negative exponent rule tells me it's equal to the reciprocal of this 1 over the power with a positive exponent so it's equal to 1 over 5 squared which is 1 over 25. and once again this one when you're saying two things are equal easily verifiable with your calculator what value of m makes this equation true if you know your exponent rules we should be able to figure this out pretty easily because notice on the left side of the equation i have 6 divided by 2 which is 3 and i have a power of a divided by a power of a i know i keep the base of the power and subtract the exponents i need to figure out how can that be equal to 3 a to the 5. so to make this equation true we just have to figure out what can i plug in for m so the exponent becomes 5 well what minus 3 equals 5 8. so my answer 8. 12 what's the correct solution to x minus 2 equals negative 4. so we have to use inverse operations to isolate the x so i'll add the 2 to the other side and i get negative 2. and any solving an equation question plug in your answer make sure it makes the equation true is negative 2 minus 2 equal to negative 4 yeah that's the correct answer and if you had no idea how to do this question you could literally one by one plug the possible answers into the equation and see which one makes it true if you thought for some reason 6 was the correct answer if you plug in 6 6 minus 2 is 4 not negative 4's we know that's not the correct answer 13 rearrange this formula to isolate h so i've got a equals b times h over 2. i want to isolate h i'll start by multiplying both sides of this equation by 2 to get rid of the fraction so i've got 2a equals these twos cancel bh and then i want to isolate h so i'll divide both sides by b because right now that h is being multiplied by b inverse of multiplying my b is dividing by b the b's cancel and we have 2a over b equals h so i know h is equal to 2a over b which is right there d 14. solve the equation for x when you have fraction equals fraction you can cross multiply cross multiplying tells you to write the product of those two terms on one side of the equal sign and the product of these two terms on the other side of the equal so i'll write 4 times 4x minus 1 on the left and 6 times 3x plus 2 on the right and then i'll solve this equation for x i'd have to do my multiplying so i'll distribute the 4 into the binomial and distribute the 6 into this binomial do all my multiplications and solve from there so i'd have 16x minus 4 equals 18x plus 12. get the terms with an x on one side and the constant terms on the other i'll bring the constant terms to the left and the variable terms to the right and collect my like terms negative sixteen equals two x divide both sides by two negative sixteen divided by two is negative eight so my correct answer negative eight and once again when solving an equation take the time plug it back in for the variable make sure it makes left side equal to right side 15 says expand and simplify so expand means do the multiplications and get rid of the brackets so i'll distribute the x into this binomial and the negative 4 into this one make sure you distribute negative 4 not positive 4. that way it gets rid of the brackets all in one step so my first product x times x is x squared x times negative 2 is negative 2x negative 4 times x is negative 4x and negative 4 times 1 is negative 4. collect the like terms my only like terms are the two terms in the middle negative 2x minus 4x is negative 6x so d is my answer y equals 66 x represents a direct variation a partial variation both direct and partial or neither so this is a linear relationship because the exponent on the x is one and since y and x vary directly with each other there's no constant added we call that a direct variation 17 on the other hand there is a constant added after the x term therefore it's a partial variation primary source of data means data you are collecting yourself so let's see which of these is someone collecting data themselves a looking up leading scorers on nhl.com well nhl.com has a staff that collected those stats so you're not collecting those stats yourself getting information about your stocks from the newspaper once again the newspaper retrieved that data not you so that's not primary c you're conducting a survey of students in your class to determine the most popular type of music that's definitely primary because you're asking the questions you're collecting the data in part d same as a and b someone else has collected the data you're just referencing the data they've collected so a b and d are secondary c is primary 19 estimating the values beyond the known data for a relation that's called extrapolation when you're estimating a value within the set of data that you have that's called interpolation and we use the line of best fit to make those extrapolations or interpolations but this definition is about extrapolation okay we have a distance time graph here in which section is rena moving the fastest well speed is a rate of change and rate of change is represented by the slope of the lines we're looking what section of this line is the steepest now by just looking at this i need to decide between p and r i think those segments are the steepest let's figure out which one is steeper section p what's the slope it would be rise 1.5 run to slope of p equals rise over run so 1.5 over 2 which is 0.75 and we're in kilometers per minute and now let me calculate the slope of r now it's going to be a negative slope because the line is going down to the right but the absolute value tells me the speed now the slope of r is down 2 over 2. and it's a little bit confusing in these calculations because the y scale is by 0.5 so the x scale is by ones but we can you can figure it out so rise negative two run two that's a slope of negative one kilometers per minute right the negative meaning towards home in this case right the distance from home is decreasing you can also think of that as something called displacement and the one is telling us the speed one's bigger than .75 which means that r is a steeper slope which means section r is when she's moving the fastest 21 the falling graph show person's distance from home we could think of that once again as displacement which graph shows an acceleration away from home okay and it's away from home so that tells me that it's going to be going up to the right she's going to be starting close to home and finishing further away from home so that narrows it down to either this graph or this graph now i need to look at which one is showing an acceleration away from home which means the slope is getting steeper as we move to the right this one as we move to the right the slope starts off pretty shallow but it keeps getting steeper and steeper and steeper so moving away at an accelerating rate so part c a would be moving towards home at a decelerating rate b moving away at a constant rate and d since the distance isn't changing she's not moving in that scenario oh and part d asks us that graph d the person's not moving no movement 23 super hot fire's earnings vary directly with the number of rap battles he wins if he earns 250 for winning five battles what is the constant of variation this question tells us it's a direct variation so y equals m x where y dependent variable would be his earnings x independent variable numbers of rap battles he wins and m is the constant of variation or the slope or the rate of change whatever you want to call it now it tells us his earnings for a specific number of rap battle wins so if we want to calculate m we can plug in what we have 250 equals m times five then isolate m by dividing both sides by five 250 divided by five is fifty so i know if he made two hundred fifty dollars for winning five battles that must mean he earns fifty dollars per win so our answer b 24 calculate the slope of this line well i know slope from a line is rise over run or if you want to do coordinate systems you can do y 2 minus y1 over x2 minus x1 but when we have the graph here rise over run is probably easy enough let's calculate the slope from e to f to get from e to f i would have to go down four units that means the rise is negative four and then to the right 5 units that means the run is positive 5. now it wouldn't matter if instead we had done from f to e if we had done from f to e what would have happened we would have had to go up 4 units which would mean the rise is positive 4 but then left 5 units which means the run is negative 5. negative 4 divided by 5 and 4 divided by negative 5 are both negative 0.8 it's the same answer it doesn't matter whether your negative is in the numerator or denominator same answer i can find the equivalent value here negative four over five my answer to this question b okay this equation is given to us in the form y equals mx plus b it just wants to be able to make sure that you can pull out which of those numbers is the slope which one's the y-intercept slope is m y-intercept is b so our slope is negative 3 our y-intercept is 4. d 26 what is the slope of the line that passes through these two points okay well remember slope is change in y over change in x which we can calculate by doing the second y minus the first y divided by the second x minus the first x pick one of these to be your first point logically i'll pick a this is my first point every point has an x and a y i'll label that x one and y one because that's the first point it's my second point so x two y two plug into our formula eight minus negative six be careful when subtracting a negative divided by 7 minus 2 i get 14 over 5 and i see that right here hey which lines are parallel well parallel lines have the exact same slope i don't care about the y-intercepts we can forget about the y-intercept so let's just look for a pair of lines that have the same slope slope of two slope of three those are not parallel slope of five slope of five okay those are parallel b let me just check the other ones not parallel not parallel b what is the slope of a line perpendicular to y equals 2 3 x plus 5. well here's the slope of that line two-thirds a perpendicular slope is the negative reciprocal of that which means i need to flip it and change its sign i need to make it negative and we can verify if two slopes are perpendicular by checking if the product of the two slopes is negative one let's check is two-thirds times negative three over two is that equal to negative one if it is then the slopes are perpendicular if i multiply these fractions i get negative six over 6. does that equal negative 1 of course it does we verify that they're perpendicular so negative 3 over 2 is the correct answer p 29 what is the slope of a vertical line vertical lines have an undefined slope c if it was horizontal would have a slope of zero what is the equation of the following line okay well here's the horizontal line the equation of any horizontal line is y equals whatever the y-intercept is because that's the y-coordinate of every single point on this line what are the y-coordinates of all these points the y-coordinates of all those points are 5. that's why we define the equation of the line as y equals 5 y it's always 5 on that line there's my answer a 31 says use first differences to determine which table values represents a linear relationship if it's a linear relationship that would mean change in y over change in x for each pair of points is the same it's a constant of variation so all we have to do to verify that is first let's verify that the x values are going up by a constant amount for each table in each table the x values are going up by one each time that's true for all tables so all we need to verify now is which of these tables have the y values also going up by a constant amount the first table goes up by 2 by 2 again but then up by 4. so the y values are not changing at a constant rate so that one does not represent a linear relationship these ones the y values go up by one then three then five okay also not a linear relationship this one by four seven four no this one finally three three three the slope of this one is three we did change in y over change in x between any pair of points we would get a slope of three so this one's a linear relationship it's d thirty-two calculate the perimeter and area of the following okay let me start by calculating the perimeter so perimeter is adding up the side lengths of all the sides on the outside of this object if this is 12 this is 12. if this is 7 this is 7 i need the inside one here well i know if this is 18 this one would be 18 minus 2 minus 3 so 18 minus 5 is 13. find the perimeter i would add up all the sides i'll start with this one and then move clockwise around so it would be 2 plus 7 plus 13 plus 7 plus 3 plus 12 plus 18 plus 12. i would have a perimeter of 74. okay well part c is the only one that has that perimeter that's going to be my answer let me verify the area as well just in case i made a mistake with the perimeter that would let me know that i have a mistake somewhere okay but i want the area of this object uh it looks like it's almost a complete rectangle it just has this part cut out of it so if i want the area i can do the area of this whole rectangle minus the area of the blue rectangle so the area of the full rectangle will be 12 times 18. minus the area of that blue rectangle which i shaded is 7 times 13 right area of rectangle's length times what so i have 216 minus 7 times 13 70 2191 that gives me an area of 125 which yes is what we have here so i've just verified okay 32 i'm very confident that's the correct answer based on the diagram below which of the following statements is correct angle abc is 40. well angle abc if i went from a to b to c it would be this angle in here uh that angle is not 40. i know that angle supplementary with 110 degree angle right angles on a straight line add to 180 so i know this is 70 degrees right here a b is equal to bc this side and this side are equal no i don't think those sides are equal because what i have here let me just fill in this angle right here this angle is 40. i know that for a couple different reasons exterior angle theorem tells me that that angle and that angle have to add to 110 but also i know three angles in a triangle have to add to 18. so i can tell from this triangle that it's isosceles so ac and bc are equal a b and bc are not equal so so far those two are not true oh triangle abc is isosceles yes it has two equivalent angles so this is an isosceles triangle so the answer to this c based on the intersecting lines which of the following statements is not true angle a equals angle c yeah those are opposite angles so those are equal so that is true angle d equals angle b yep those are opposite angles that's also true angle a plus angle b equals 90. well a and b are supplementary they form a straight line so i know they add to 180 so this one is not true let me just check with part d angle d and angle a add to 180 yes those form a straight line they're supplementary they do add to 180. so c is the only one that's not correct that's my answer 35 so i've got two parallel lines cut by a transversal which of the following statements is not true angle d and f equal 180. yep d and f are co-interior angles they form the c pattern which tells me that yes they do add to 180 degrees so this one's true b and h are equal b and h b and h are not equal because i know d and h are equal and d and b are not equal so b and h i don't think those are equal to each other let me just verify that the other ones are true c and f are equal yep they form a z pattern so c and f alternating angles are equal and g and e equal 180 g and e yes they form a straight line so they're supplementary so yes they do add to 180. so b is the only one that is not true 36 our last multiple choice question g and h are complementary okay so complementary means they add to 90 degrees they form a right angle it gives me an expression for each of the angles and it says what is the measure of each angle well i know they add to 90 so i could set 3x plus 6 plus 2x minus 11. i could set those equal to 90. i could solve for x plug both my answers into each angle and get the measure of each angle so let me collect my like terms 3x plus 2x is 5x 6 minus 11 is negative 5 equal to 90. isolate the variable term by adding 5 to the other side divide the 5 to the other side and i get x is 19. if x is 19 i can now sub for angle g and h angle g is 3 times x which is 19 plus 6 and angle h is 2 times x which is 19 minus 11. if i do this i would get 63 degrees and if i do this i would get 27 degrees let's see is there a 63 and 27 yes angle g and h 63 and 27 hey okay that's multiple choice done [Music] we are now on to the short answer section of the exam section two is questions about algebra number 37 says simplify the following expressions using exponent laws and evaluate where possible if we look at part a i have a product of two terms we need to look at the factors of each of the two terms and see if there are any pairs that when i multiply them together it will simplify i'll start by looking at the constant factors of each of the terms they're the coefficients of the two of them 2 and 5 they're being multiplied and 2 times 5 is 10. then i'll look at the variable factors of the terms and find the ones that have the same base i have an a to the 6 being multiplied by an a cubed the product rule for powers tells me that when i'm multiplying powers with the same base i keep the base so in this case a and add the exponents 6 plus 3 is 9. and then i also notice that both of the terms have a factor of b they're being multiplied together so i'll keep the base b and add their exponents 1 plus 3 is 4. now that'll be my final answer for this i can't evaluate it because i don't know the values of a and b they're variables but for part b i should be able to evaluate this one let me simplify it first using exponent laws though because so if i look at the first term i have a 3 squared to the power of 5. the power of a power rule tells me to keep the base of the power 3 and multiply the exponents 2 times 5 is 10. that's being divided by a 3 to the power of 7 and when dividing powers with the same base you keep the base the same and subtract the exponents 10 minus 7 is 3. and now i need to evaluate this now that i've fully simplified it using exponent rules if i evaluate this 3 cubed that means 3 times 3 times 3 which is 27. now for this question if you're writing this exam you could definitely check and make sure this is right you could just type the original expression on your calculator and it'll tell you that it's equal to 27. part c i have 5x to the 8y to the 8 divided by 15x cubed y to the 5. this is similar to part a but instead of the terms being multiplied they're being divided so what i'm going to do is look for factors in the numerator and denominator that simplify when dividing them i'll start by looking at the constant factors 5 divided by 15. think of it as a normal fraction 5 over 15 see if you can reduce it 5 goes into 5 one time and into 15 three times so i can reduce 5 over 15 to 1 over 3. now look for factors that have the same variables so i have factors of x and factors of y being divided if i do the quotient rule of exponents it tells me that when dividing powers with the same base you keep the base and subtract the exponents so when i divide the factors of x i'm going to keep the base of x and subtract the exponents eight minus three is five and i divide the powers of y i'll keep the base of y and subtract the exponents eight minus five is three notice when you're doing these divisions always put the quotient into the numerator of your fraction and then if the exponents on any of your factors were negative you would then take that factor and move it to the denominator to make the exponent positive we'll see an example of that in the next question but part c there's nothing else we can do to that i mean we wouldn't write a factor of 1 so i'll just rewrite this as x to the 5 y cubed over 3. part d part d once again i have two terms being divided i'll start by dividing the constant factors four divided by two is two and you could think of that as two over one if you want and then i have two factors of x being divided so when i divide them i would keep the base of x and subtract the exponents 2 minus 4 is negative 2. you should never leave an answer with a negative exponent in it so you take the factor that has the negative exponent so the x to the negative 2 and i'll move that to the denominator which will change the sign of the exponent so this would become 2 over 1x squared which i'll just write as x squared so that would be the final answer to part d part e whenever you have a question that is a quotient you want to make sure you can simplify the numerator and denominator as much as possible before you try and do any dividing so let me simplify the numerator in the numerator i have a power where the base is a product of two factors a 2 and an x cubed so i have to put the exponent of 3 onto both factors of the base i have to cube the 2 and i have to cube the x cubed in the denominator i've got two terms being multiplied i'll multiply the constant factors four times two is eight and i'll multiply the variable factors the x squared and the x to the four because they have the same base i can keep the base and add the exponents 2 plus 4 is 6. in the numerator i have a 2 cubed which is 8 and an x to the 3 to the 3 the power of a power rule tells me to keep the base and multiply the exponents that's eight x to the nine so i have eight x to the nine over eight x to the sixth now that the numerator and denominator are fully simplified i can do my divisions if i divide the constant factors eight divided by eight is one and x to the nine divided by x to the sixth i would keep the base of x and subtract the exponents nine minus six is three so i have one x to the three which we would just write as x to the power of three those are all the exponent rule questions let's now move on to question 38 which asks us to expand and simplify the following so part a wants me to do this monomial 3 multiplied by the binomial of x plus 2. to do that multiplication correctly i have to multiply the x and the 2 by the 3 that's out front so i'll find the product of 3 times x which is 3x plus the product of 3 times 2 which is 6. part b i have two binomials being added together anytime you're adding polynomials you can just drop the brackets and collect the like terms like terms are terms that have the same variables with the same exponents so i have a 2x to the 1 plus another 2x to the 1. i'll just write those beside each other so you can see they're like terms and i have two constants of 5 and a negative 7. i'll be able to collect those together as well those are also like terms 2x plus 2x when collecting like terms you just add the coefficients so 2 plus 2 is 4 and you keep the variable part of the term the same so 2x plus 2x is 4x and positive 5 minus 7 is negative 2. part c i have a 4x times the binomial 3x minus 5 minus 7 times the binomial x squared plus 2x to expand and simplify this i need to do the multiplications and then collect all the like terms so i'll distribute the 4x to the 3x and the negative 5 and i'll distribute the negative 7 to the x squared and the 2x when i do those multiplications the brackets will be gone and then i can start collecting my like terms together so let me find my first product 4x times 3x the constant factors get multiplied 4 times 3 is 12 and the factors of x get multiplied x times x that's an x to the 1 times an x to the one so i keep the base of x and add the exponents 1 plus 1 is 2. now i'll do 4x times negative 5 4x times negative 5 will give me negative 20 x now let's distribute the negative 7 to the x squared and the 2x negative 7 times x squared is just negative 7x squared and negative 7 times 2x would be negative 14 x i'll look for like terms i have a 12x squared and a negative 7x squared those are like terms because they have the same variable factors with the same exponents they both have an x squared i have a negative 20x and a negative 14x those both have an x to the one so those are also like terms when i collect these together 12x squared minus 7x squared i just subtract the coefficients 12 minus 7 is 5 and i keep the variable part the same x squared and i have negative 20x minus 14. negative 20 minus 14 is negative 34. and then keep the x the same part d i have nested brackets you're going to want to start with the innermost set of brackets and then work your way out so i'm going to distribute the 4 to the x and the 3 to get rid of that innermost set of brackets so i will leave the 2 on the outside and then i have a 2x that will just stay plus now i'm going to find the product of 4 times x and 4 times 3. 4 times x is just 4x and 4 times 3 is positive 12. so i've gotten rid of the innermost set of brackets i notice inside those square brackets i drew i have a 2x plus 4x those are like terms i could collect those together to be 6x and now we'll distribute the 2 to both the 6x and the 12. 2 times 6x is 12x and 2 times 12 is 24. if you notice for the last two questions we just did we were simplifying expressions we just had an expression and we wanted to write it in an equivalent but simplified form so each line i wrote started with an equal sign showing that it's equivalent to the line above that's an important thing to make sure you're doing to clearly communicate your solution so that's something that i look for when i'm marking a student's work i want to make sure they're accurately communicating what they're doing their work is organized and on each line it starts with an equal sign we're now going to move on to solving equations let's move on to question 39 where it says solve the following equations and it even puts a reminder don't forget you can check your solution so instead of having an expression that needs to be simplified we now have two expressions that it says are equal to each other and our goal is to figure out what value of the variable makes the two expressions equal to each other and we'll follow some algebra rules to help us isolate the variable to figure out the answer so for part a 3x minus 17 equals 13. that means 3 times something minus 17 is equal to 13. we'll start by isolating the term that has the x we'll isolate the 3x right now 17 is being subtracted from 3x i can move this term to the other side by doing the inverse operation so the inverse meaning opposite of subtracting 17 would be adding 17. so what i have is 3x equals 13 plus 17 and 13 plus 17 is 30. now i have 3 times something is 30. i'm sure you know at this point the answer is going to be 10 because 3 times 10 is 30 but how do we show that algebraically we want to isolate the x and currently it is being multiplied by 3 so we can move that factor of 3 to the other side by doing the inverse of multiplying by 3 which is dividing by 3. so i would have x equals 30 divided by 3 and 30 divided by 3 is 10. we can check and make sure 10 is the correct answer doing a quick left side right side check the left side of the equation is 3x minus 17. and the right side is 13. if i plug my answer of 10 in for the variable x it should make the left side and right side the same so let's check i'll plug in 10 for x i get 30 minus 17 on the left side which is 13. notice that's the same as the right side so we have the correct answer to the equation one other thing i want to add if we look at the second line of my solution really what's happening between the second and third line is i'm dividing both sides of this equation by three whenever you do algebra really what you're doing is you're doing the same thing to both sides of the equation so really what's happening is i'm dividing both sides by 3 so that the factors of 3 on the left cancel and i'm left with just x on the left and a 30 divided by the 3 on the right just like the first step to this question really what happened was we added 17 to both sides of this equation it makes those terms on the left cancel out so all we're left is 3x on the left and on the right 13 plus 17. part b has a fraction we can always get rid of any fraction you have in an equation by multiplying both sides of the equation by the denominator of the fraction so let's start by doing that i'll multiply the entire left side by four and the entire right side by four we can do whatever we want as long as we do it to both sides the reason why this is useful is because on the left i have a four divided by four which is one so those cancel out and on the left i'm just left with two x plus five and on the right i have four times two which is eight i'll now isolate the term that has the x so i'll isolate two x by moving the plus five to the other side and it becomes a minus five so two x equals eight minus five so two x equals three the x is being multiplied by two so to isolate it i do the inverse of multiplying by two which is dividing by two so on the right i have x equals 3 over 2 which is 1.5 but we can leave our answer as a fraction so there's my final answer x equals 3 over 2. part c notice i have two terms that have the variable p that i'm trying to solve for so i'm going to need to eventually get all the terms that have a p to the same side of the equation before i can do that i'll have to expand the left and right side of the equation so on the left i'll distribute the 3 to the 2p and the 1 so i'll have 3 times 2p which is 6p and 3 times 1 which is positive 3 equals and on the right 5 times p is 5p and 5 times 1 is positive 5. now i need to get all the terms that have the variable to one side and all the constants to the other side so i'll bring the variable terms to the left so i have a 6p and when i bring the positive 5p from the right side across the equal sign to the left side it's going to change its sign so it's going to become a negative 5p on the right side of the equation the 5 is staying and i'm bringing the positive 3 from the left across the equal sign to the right so it's going to change from a plus 3 to a minus 3. 6p minus 5p is 1p and 5 minus 3 is 2. so there's my final answer p equals 2. and don't forget you could check that in the original equation to make sure it's true if i plugged in 2 for p on the left i would have two times two which is four plus one is five times three is fifteen so the left would be fifteen and the right if i plugged in two for p two plus one is three times five is fifteen so it makes the left and right both fifteen therefore that's the correct answer part d i have a fraction equals a fraction when you have that you can cross multiply as a shortcut or any time you have multiple fractions you can get rid of them by multiplying both sides of the equation by a common multiple of your denominators so a common multiple between 8 and 4 is 8. so i can multiply both sides by 8. on the left side of the equation i have 8 divided by 8 which is 1 so those cancel and on the right i have 8 divided by 4 which is 2. so the left side of the equation is just 3x plus 2 and the right side is 2 times 3x minus 2. i have two terms of the variable i'll need to get them on the same side but before i can do that i need to expand the right side by distributing the 2 to the 3x and the negative 2. so 2 times 3x is 6x and 2 times negative 2 is negative 4. i'll get the variable terms to the right this time and the constant terms to the left so the 2 will stay on the left side of the equal sign i'll bring the negative 4 from the right side of the equal sign across to the left so it becomes a positive 4. the 6x is staying on the right bring the positive 3x from the left to the right side of the equation so it becomes a negative 3x so i have 6 equals 3x to isolate the variable that's being multiplied by 3 i have to do the inverse of multiplying by 3 which is dividing by 3. so 6 over 3 equals x so i know x is 2. and i know you could have solved this question by cross multiplying instead let me just very quickly show you that that would get you to the same answer when you have fraction equals a fraction you can cross multiply i can do the product of 8 and 3x minus 2 on one side of the equation and the product of 4 and 3x plus 2 on the other side of the equation and then solving this equation will give us the same answer of two that we got doing it the other way probably the most efficient way to solve this equation from here would be to divide both sides by four that'll cancel those out and eight divided by 4 is 2 but i would probably see most students at this point just start distributing to expand i mean it'll work it'll get you the same answer let me just show you we would have on the left 24x minus 16. and on the right 12x plus 8. bring the variable terms to the left and the constant terms to the right i'd have 12x equals 24. 12 times what is 24. well if i do 24 divided by 12 it'll tell me that the answer is 2 same answer we got by doing the question the other way the last equation to solve part e i see three x's i'm going to need to get them all to the same side of the equation so i'll start by expanding to get rid of the brackets so that i can start moving terms around on the left i'll distribute the 3 so 3 times 2x is 6x and 3 times negative 5 is negative 15. i also have this term of negative x i'll leave there on the right a common mistake here would be to think you have to distribute the 4 but that's not what we have it's not four multiplied by three x minus seven it's four minus three x plus seven so two ways to think of this you could just think of subtracting both of those terms or you could think of it as a negative one that is multiplying into both of those terms either way you'll get the same result so this 4 is just going to stay and then negative 1 times 3x is negative 3x and negative 1 times 7 is negative 7. let's bring all the variable terms to the left so i'll leave the 6x the negative 1x bring the negative 3x over it becomes positive 3x and get all the constant terms on the right there's already a 4 and a negative 7 there and i'll bring over the negative 15 making it a positive 15. collect all the like terms on the left i have 8x and on the right i have 12. isolate the x by doing the inverse of multiplying by eight which is dividing by eight and four goes into both twelve and eight it goes into twelve three times into eight twice so it reduces to three over two notice when solving in equations the proper format is to create a new line anytime you're simplifying anything and on the new line you need a left side of the equation a right side of the equation and an equal sign between those two sides let's now do the word problems number 40 show a full algebraic solution to either part a or part e not both make sure to clearly communicate your final answer i'll answer both for you just in case you're curious as to what they would be part a says the length of a rectangle is 3 centimeters more than double the width the perimeter of the rectangle is 96 centimeters what are the dimensions the length and width of the rectangle for this question we need to solve for length and width so what i'm going to have to do is be able to write an expression for the length and width in terms of the same variable then create an equation that involves that variable and be able to solve that equation so notice the question says the length of the rectangle is three more than double the width but i don't know what the width is so i'm going to assign the width to be x and now i need to write an expression for length in terms of that same variable the length is equal to three more than double the width so it's equal to three more than double x so the length would be double x plus 3 so 2x plus 3. and i know the perimeter of the rectangle is 96 so with that information i can create my equation remember the perimeter of a rectangle would be two times the length plus two times the width in this case our length is two x plus three and our width is x and we know the perimeter so i'll make the left side of this equation 96. i now need to solve this equation for x i'll start by distributing the 2 on the right side of the equation to the 2x and the 3. 2 times 2x is 4x 2 times 3 is 6 and i have plus 2x i'll bring the constant terms to the left so i'll bring the plus 6 over becomes a minus 6. on the right i've got 4x plus 2x this simplifies to 90 equals 6x and to isolate the x i'll have to do the opposite of multiplying by 6 which is dividing by 6. so i'll have x equals 90 divided by 6 which is 15. so x is 15 which means my width is 15 centimeters and my length is 2 times 15 plus 3 which is 33 centimeters we should write a final answer as a sentence we should say the rectangle has a length of 33 centimeters and a width of 15 centimeters let me now erase this and let's try part b instead part b says in a triangle the measure of the middle angle is double the measure of the smallest angle and 15 degrees less than the measure of the biggest angle find the measure of the angles use a diagram to help so the three things i'm trying to solve for in part b are the small angle the middle angle and the big angle so i have to come up with an expression using the same variable for all three of these so let's see what the first sentence says it says the measure of the middle angle is double the measure of the smallest angle but i don't know what the smallest angle is so let me start by calling the smallest angle x and i know the middle angle is double that so it would be 2x and then let me keep reading and it also says so not only is the middle angle double the measure of the smallest but it's also 15 degrees less than the measure of the biggest angle another way of saying that would be the biggest angle is 15 degrees more than the middle angle so the biggest angle will be 2x plus 15. now i can create an equation because i know the angles in a triangle have to add to 180. so i know that x plus 2x plus 2x plus 15 has to be equal to 180. i'll keep all the variable terms on the left i'll bring the plus 15 to the other side becomes minus 15. so i have on the left 1x plus 2x plus 2x that's 5x and on the right 180 minus 15 is 165. divide both sides by 5 and 165 divided by 5 is 33 so my smallest angle is 33 my middle angle is double that which is 66 and my biggest angle is even 15 more than the middle angle so it's 81. do a quick check make sure they add to 180 i would have 33 plus 66 which is 99 plus 81 which is 180. and we should write a sentence saying the angles of the triangle are 33 degrees 66 degrees and 81 degrees and there's the final answer let's move on to our next section linear relations [Music] question 41 is about data collection and different sampling methods so you'll have to understand the definition of simple systematic stratified and non-random sampling so we'll look at each of the four scenarios and see which one matches which scenario one says you give a survey to the tallest students in your class in that scenario there's no random method used at all you're picking the people yourself and you're choosing them because they're the tallest that's a perfect example of non-random sampling so scenario one is d scenario two giving a survey to the ten students whose names were drawn from a hat so when everyone in a population gets an equal chance of being picked on each draw that is called a simple random sample so scenario two simple random sampling scenario three giving a survey to ten percent of girls and ten percent of boys in the class so in this scenario the population was divided into two groups and then within each group a simple random sample of an equal proportion were done that is stratified random sampling the key thing i noticed there is that the population was put into groups and then within each group a simple random sample was done so that's stratified scenario four giving a survey to every fourth person alphabetically in that case when you sample at an interval that's systematic random sampling so scenario b question 42 is based on the following table of values is this a direct or partial variation i see when x is 0 y is 3 so i know this is a partial variation for it to be a direct variation the initial value would have to be zero meaning when x is zero y would be zero but in this case the initial value is three when x is zero y is three so we see that's a partial variation part b says to fill in the missing information in the table so if i look at the x values it goes 0 1 2 3 4 the y values go 3 7 11 15 and then i'm not seeing the value of y that goes with the x value of 4. if we look at the pattern the x values are going up by one and the y values are going up by four so the next y value must be 19. but then notice it skips from 19 to 31. so it went up by four how many times 19 to 23 would be going up by 4 once to 27 twice to 31 three times so it went up by four three times so i must add three to that x value which is seven so it skipped over the point five twenty three six twenty seven and it went to the point seven thirty one part c says what is the constant of variation which we call the slope or the rate of change we use the letter m to calculate m we need to calculate what's the change in y divided by the change in x and we can do that in an organized way by doing y2 minus y1 over x2 minus x1 so we have to pick a point to be our first point and a point to be our second point we might as well just use the first two points i'll make the first row my first point so that's my x1 y1 and the second row is my second point so that's my second x coordinate and my second y coordinate if i follow the formula it tells me to find the change in y so i would do 7 minus 3 divided by the change in x x two minus x one is one minus zero so i get four over one which is four my m value is four what's the initial value that means when x is zero what's y when x is zero y is three and we have a variable we use for that initial value we call it b so i can say b equals 3. write an equation in the form y equals mx plus b just take your answer for m and b and plug them into the equation so i would have y equals 4x plus 3. this equation defines the relationship between the x and y value of any point on the line in terms of its slope and y intercept so we could actually check and make sure these answers from part b are correct assuming we have the correct equation showing the relationship between x and y if i want to check and make sure 19 is correct when x is 4 well let's check what do i get for y if i plug 4 in for x i get 16 plus 3 which is 19. so that's good but what do i get for x if i plugged in 31 for y 31 plus 4x plus 3 let's subtract the three over i would get 28 equals four x divide the four i get seven so that's correct as well let's move on to the next question which asks us to calculate slopes of lines so remember slope we use the variable m to represent that and we calculate it by doing change in y over change in x which graphically speaking is rise over run or algebraically if we wanted to calculate it we could find the coordinates of the points and do y2 minus y1 over x2 minus x1 it's up to you just whatever you're doing make sure you pay attention to the scale of your graph and make sure you know what your scale is going up by in this case our scale is going by ones so when i move down or up or left or right a unit on my grid it is by one so it should be easy to calculate this one i'll show you for the first line segment um from a to b how we could calculate the slope both ways just by counting rise and run and by doing it algebraically using our change in y over change in x so if i were to just count rise and run i have to pick what point i want to start at i can either start at a or b doesn't matter and then go to the next one so to get from a to b i would have to go down one two three four five six units so that would mean that my rise is negative six from going down that's a negative rise and then i would have to go to the right two units going to the right is a positive run so that would give me a run of two and negative six divided by two it simplifies to negative three so my slope is negative three we could have gotten that same answer by writing the coordinates of both of the points and then assigning one of them to be the first point i'll just make a my first point so that's my first x first y point b i'll make my second point so that's my second x and my second y and then if i follow the formula y two minus y one over x two minus x one i would get zero minus six divided by six minus four that gives me negative 6 over 2 which once again is negative 3. the same answer just approaching it in a different way cd let's just do it using rise over run so to get from c to d i would have to go up 1 two three four units and when you're going up that's a positive rise so my rise is positive four and my run from there i would have to go to the right one two three four five six so my run is six and when going to the right it's positive four over six simplifies to two over three one thing i want you to notice is that a line segment that's going down to the right is going to have a negative slope and a line segment that's going up to the right is going to have a positive slope so make sure the sign of your final answers always makes sense based on what you're seeing in the graph 44 use the graph of this line to answer the following questions what's the y-intercept of line so i would look and see where does it cross the y-axis and i actually don't see a scale here so i'll assume it's going by ones but here's the y-intercept it's at the point 0 negative 2. so the y-intercept we use b as the variable for the y-intercept i'll say b is negative 2. the slope of the line i'll just have to count the rise and the run to get between any pair of points on this line i'll do the two points it has labeled for me to get between those two points i would have to go up three and right two up is a positive rise so positive three and right is a positive run so positive two so the equation of this line in the form y equals mx plus b is y equals 3 over 2 x minus 2. question 45 wants us to graph the following lines notice for part a the line is already in the format y equals mx plus b the b value is negative 4 and we always start by plotting the y intercept which is the b value of negative 4. so i'll start by plotting the point on the y axis at negative 4. that's always the first thing you do for these questions and then from that point you use the slope which is the m value which is the coefficient of the x 1 over 2 in this case to plot more points the numerator of your slope is your rise and the denominator is your run so from that y-intercept i have the rise 1 which means up 1 and run 2 which means right 2. and then just keep rising 1 and running 2 to plot more points fill the grid with as many points as you can that's all the points i can plot to the right of the y intercept i can plot points to the left of the y intercept by instead of going up 1 and right two do the opposite go down one and left two and notice that will give you more points that fall on the same line why does that work well because one over two is equivalent to negative one over negative two they both have a value of 0.5 they're equivalent to each other so you can either go up 1 right 2 or down 1 left 2 and it gives you points on the same line and when drawing your line through the points make sure you put arrows on both sides showing that it is a continuous line that continues forever let's do part b it's also already in the format y equals mx plus b my b value is the constant at the end it's two so my y intercept is going to be at two and my slope my m value is the coefficient of the x now this time it's negative three but it'll help for me to think of that as a fraction negative three over one so that i can see what the rise and the run are any whole number you can write over one so my rise is negative three that means from my y-intercept that i plotted i'm going to go down three one two three and then run one which means right one and then fill my grid with points keep going down three right one and then to get points to the left of the y intercept do the opposite of down three right one which would be up three left one connect them with a line and make sure you have arrows on both ends of your line my next two equations are special examples of lines part c is a horizontal line it's the horizontal line y equals four so that means the y value of my line is always four the only way the y value of the line could be always four is if it is a horizontal line that crosses through 4 on the y axis notice that this line the y value of any point on this line would be 4. that's why we define the equation of this line as y equals 4. part d the equation is x equals negative three so the x value is always negative three the only type of line which would have a constant x value that is not changing would be a vertical line so it'll be the vertical line that passes through negative 3 on the x-axis so x equals a number is a vertical line and y equals a number is a horizontal line and the number will actually tell you where you'll cross the x or the y-axis 46 this is the first time where it doesn't give us a graph it just gives us two points on the line and asks us to find the equation of the line our end goal for this is to write the equation in the format y equals mx plus b so we'll have to solve for m and b let's start by solving for the slope of our line m and we can use our formula y2 minus y1 over x2 minus x1 pick either a or b to be your first point it doesn't matter i'll pick a to be the first point each point has an x and a y so i'll call those coordinates x 1 y 1 and point b i'll call those coordinates x 2 y 2. if i follow the formula for slope i would have 2 minus 8 divided by negative 2 minus negative 5. this would give me negative 6 over negative 2 minus negative five is negative two plus five which is three so my slope is negative six over three which is negative two so my slope is negative two to solve for b into y equals mx plus b i need to sum in the slope i just solved for which is negative two and any point x y that is on my line i know two points that are on the line you could pick either point a or point b to be your x and y values i'll pick point b the y value is two the x value is negative two and then we will solve for the only unknown b so on the right i have a product of negative two and negative two that's 4. to isolate b i'll subtract the 4 to the other side so i get b is negative 2. so my final equation will always leave y and x but sub n m and b and in this case m and b are both negative two so it'll be negative two x plus negative two which i'll write as minus two forty-seven is very similar question but this time it gives us the slope tells us the m value is 2 and it tells us what point it goes through so into y equals mx plus b i can sub in my point for x and y my slope for m and solve for b and then i can write the equation y is negative three m is two x is four solving this for b i would move the eight to the other side making it minus eight and i get b is negative eleven so my final answer will be y equals 2x minus 11. question 48 find the equation of the line that is parallel to the line 4x plus 5y plus 2 equals 0 and goes through the point 5 negative 3. this is very similar to the previous question it gives us the slope and a point it's just it gives us the slope in a little bit of a tricky way it tells us it's parallel to this line which means it has the same slope as that line but i'll need to rearrange the equation of that line it gives us into the format y equals mx plus b to be able to see what the slope is so i need to isolate the y i'll move the 4x over making it a negative 4x move the positive 2 over making it a minus 2. i need to isolate the y it's currently being multiplied by five so to the other side i need to divide by five i need to divide all the terms by five so be negative four over five x minus two over five so i can now see that the slope of this line is negative four over five so the slope is negative four over five so i know the m value the only thing i don't know is b the y intercept so i can plug in the slope that i know negative four over five and the point 5 negative 3 as my x y and then solve for b so y is negative 3 m is negative 4 over 5. x is 5 and then i don't know b this will simplify nicely because i have a five divided by five that's one those cancel out so i have negative three equals negative four plus b move the negative four over becomes positive four and i get b equals 1. so my final equation is y equals negative 4 over 5 x plus 1. question 49 says find the x and y intercepts of this line 4x plus 5y equals 20. i'll start by finding the x-intercept i know that wherever the function crosses the x-axis the y-coordinate is going to be 0. so all i have to do is set the y-value to 0 by subbing in 0 for y and solving the equation so i have 4x plus 5 times 0 that's 4x plus 0 equals 20. divide the 4 to the other side and i get x is 5. so my x-intercept is going to be at the point 5 0 to find the y-intercept i know wherever the line crosses the y-axis the x-coordinate is going to be 0. so 7 0 for x and solve this equation for y 4 times 0 is 0 that's gone i have 5y equals 20. divide the 5 to the other side and i have y equals 4. so the y-intercept is at the point 0 4. to find the x-intercept set y to zero and solve find the y-intercept set x to zero and solve here is our last linear relationship question data has been collected from basketball player the following table shows a number of successful jump shots michael jensorden made at various distances from the basket identify the independent and dependent variables in a table of values you're always going to find your independent variable on the left that's our x and the dependent variable which we usually call y will be on the right column of your table so our independent variable is the distance from the basket that's the one that we control and the dependent variable is shots made and that's the one that should be affected by your independent variable the distance so the number of shots you make should depend on the distance you are from the basket part b asks us to graph the relationship make sure you put your independent variable on the x-axis so the x-axis will be my distance from the basket and my y-axis is going to be the shots made and now i need to pick a scale to go up by for my independent variable for distance i need to go as high as 11. so if i look at how many spots i have here i think i have 22 units along the x-axis i could go up by 0.5 and make 11 fit so i'll label every other spot going up by ones and on the y-axis i need to go as high as at least 22. it looks like i will have to go up by ones to make this fit so i'll label every other spot going up by twos and now i'll have to plot each point that i get from the data that was collected from three meters 22 shots were made so i'll go to three on the x-axis and then go up to 22 on the y-axis and plot a point from 4 meters from the basket i don't know how many shots were made there's probably going to be a question about that later from 5 meters from the basket 17 were made so my x value is 5 my y value is 17. so i'll plot a point at 5 17. from 6 meters from the basket 14 shots were made so i'll go to the point 6 14 7 meters 15 baskets were made eight meters ten baskets were made nine meters four baskets were made 10 meters three baskets were made and 11 meters only one basket was made part c asked me to draw a line or curve of best fit so it looks like a fairly linear relationship so i'll draw a line of best fit when drawing a line of best fit you're going to want to make sure you go through as many points as possible keeping roughly an equal number of points above and below the line and make sure your line shows the trend in the data fairly well so i think my line of best fit will look like this notice i've got a few points below the line a few points above the line it shows the trend and the data i think that's a fairly accurate line of best fit and part d says is the relationship linear or nonlinear well i decided that it was linear because a line seemed to represent the data well and it says predict the number of shots you would make if you were four meters from the basket four meters from the basket to make a prediction i'll have to figure out where is my line of best fit at 4 meters it looks like it's right about there pretty close to a y value of 21. so my trend line predicts that there would be 21 baskets made second part of this question says was that interpolation or extrapolation this was interpolation because i estimated for a distance of 4 meters and that's within the range of the data i have i have data that goes from 3 meters to 11 meters so any estimation between 3 and 11 would be interpolation but any estimation for a value beyond 3 and 11 so either less than 3 or bigger than 11 that would be extrapolation so this one was interpolation [Music] last section of the exam is geometry question 51 says find the volume of either of those shapes volume is always area based times height for any type of prism for rectangular prism the volume is just length times width times height based on your perspective you could assign any of those three as your length width or height i'll say my length is nine my width is eight and my height is six so that equals 432 centimeters cubed when doing volume you'll always get units cubed now we don't have to do part b because it says do a or b not both but i'll just show you what the answer would be the volume of a triangular base prism the formula is half blh where b is the base of the triangle l is the height of the triangle and h is the height of the prism so if i plug into this formula it would be half times 4 times 3 times 10 that would give me 60 centimeters cubed 52 says find the surface area of either a or b not both i'll show you both for a sphere the surface area formula is 4 pi r squared in this case in the diagram it looks like it's labeling the diameter to be 8 which means that the radius is half of that which is 4. let me get an approximate answer by my calculator doing the product of 4 pi and 4 squared i get about 201.1 if i round to 1 decimal place for a cone the surface area is pi r s plus pi r squared pi r s gives us the lateral surface and pi r squared gives us the area of the base of the cone now for this one i have the radius the radius looks like it is 10 and i have the height of the cone but i don't have s which is the slant of the cone so to solve for s notice the right triangle inside the cone i can use pythagorean theorem so let me off to the side just show my work s squared would equal 10 squared plus 24 squared so s squared equals 676 s would be the square root of that which is 26. so i now know the slant height is 26 centimeters so into my formula i can input the radius of 10 and the slant height of 26 and i'll type this on my calculator to get an approximate value and i get about 11 31. centimeters squared notice surface area is unit squared volume is units cubed 53 once again asked us to do either a or b not both part a there's a few ways we could do this the most efficient way would be to know that two interior angles of a triangle are going to be equal to the opposite exterior angle of the triangle so i know that x plus 68 is going to be equal to 130 subtract the 68 to the other side and i figure out x equals 62. so x is 62 and i should say degrees if i want to solve for y notice that 68 and y form a straight line so those angles are supplementary which means they add to 180 degrees therefore 68 plus y equals 180 isolate y by subtracting the 68 and i get y equals 112 degrees and the last angle i don't know is angle z it forms a straight line with x so i know that z and x add to 180 and remember we know the answer for x x is 62. subtract the 62 to the other side and i get 118 degrees let me move on to part b if i focus on the top triangle i have two of the three angles i know that the three angles in a triangle add to 180 so 65 plus 50 plus y equals 180 65 plus 50 is 115 y would equal 180 minus 115 so y is equal to 65 degrees angle y and z are opposite angles and i know that opposite angles are equal to each other so z is also equal to 65 degrees and then lastly i can solve for x if i look at the bottom triangle i know all three angles have to add to 180 so 74 plus 65 plus x has to equal 180 139 plus x equals 180 x equals 180 minus 139 x equals 41. 54 once again says do a or b i'll show you the answers for both in part a i have the two angles inside of a c pattern that's called the co-interior angle theorem i know that angles inside a c pattern have to add up to 180. so 11x plus 10 plus 10x minus 40 equals 180. 11x plus 10x is 21x 10 minus 40 is negative 30. let me move that negative 30 to the other side 180 plus 30 is 210. divide the 21 210 divided by 21 is 10. so x equals 10. part b we also have to solve for x and i know that two interior angles of a triangle are equal to the opposite exterior angle so i can set up my equation on the left if i collect like terms i have 4x plus 22 equals 5x i'll bring the variable terms to the right so i have 5x minus 4x on the right so 22 equals x we're almost done last question solve for the missing side of each triangle they're both right triangles so i know pythagorean theorem holds true for these pythagorean theorem tells me that the sum of the squares of the shorter two sides equals the square of the longer side and you may have seen that written as a squared plus b squared equals c squared c is your hypotenuse which is always across from the right angle and it's always the longest side of your right triangle so here's c the legs the shorter two sides doesn't matter which one you call a and which one you call b if i plug into my formula 36 squared plus b squared equals 39 squared i need to rearrange to isolate b so i'll have b squared equals 39 squared minus 36 squared 39 squared minus 36 squared is 225. to isolate b it's currently being squared the inverse of squaring is square rooting so i must square root 225 to solve for b and the square root of 225 is 15. part b i know the shorter two sides i know the legs a and b i don't know the side across from the right angle the hypotenuse which we'll call c if i plug into pythagorean theorem it would be 32 squared plus 26 squared equals c squared 32 squared plus 26 squared is 1 700. to isolate c it's currently being squared so i have to do the inverse of squaring which is square rooting and the square root of 1700 is about 41.2 in this case we know units so i'll write centimeters all right that's it for the exam hopefully this in some way has helped you either prepare for your exam review stuff from when you took grade 9 or helped you look ahead and preview what's in a grade 9 course