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Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Lesson 4: Topic D: Systems of linear equations and their solutions

Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60

Sal solves the system of equations 6x - 6y = -24 and -5x - 5y = -60 using elimination. Created by Sal Khan.

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  • leafers ultimate style avatar for user Sebastian Guo
    if you multiply one equation by 1/6, shouldn't you also multiply the other equation by 1/6, not 1/5 so they are equal?
    (12 votes)
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  • piceratops seedling style avatar for user Étoile
    I don't understand why you have to multiply the equations by " 1/6 & 1/5 "..
    If i have a problem like " 9a - b = 15 , 3a + b = 5 " how do i solve this with elimination..
    (6 votes)
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    • aqualine ultimate style avatar for user George, Aaron
      You don't always do elimination. You can divide both sides by the same thing or subtraction to equal out the equation to the maximum level of simplification. Then, great! You have a fully simplified equation, to find out what is greater, or finally solve for the unkwown variable.-- Hope it helped,


      GEORA190
      (1 vote)
  • hopper cool style avatar for user Deven
    How would you know what number to multiply the equations by? In the video its pretty obvious but for some reason, I can't get it on my own. For example: 4x-3y=-26 5x-4y=-35
    Or: -3x+5y=29 -3x+6y=39
    (3 votes)
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    • leaf blue style avatar for user Chamuel Patino
      4x-3y=-26
      5x-4y=-35
      You want eliminate either the x or the y. Let's say we eliminate the x first, we have to make the top (4x) and (5x) equal to each other. Multiply the top by 5 and the bottom by 4. Now both are (20x) and you can subtract to cancel them out. Remember that whatever number you multiply by is distributed to the whole equation.
      (8 votes)
  • mr pink red style avatar for user ysysteve
    How do I solve this :" if a - b = 2, a - c = 1/2, then what is the answer for (b - c)^3 - 3(b - c) + 9/4"
    (4 votes)
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    • leaf green style avatar for user Agent Smith
      Let equation (1) be a - b = 2 and equation (2) be a - c = 1/2
      equation (2) - equation (1) = (a - c) - (a - b) = (1/2) - 2 = b - c = -3/2
      we now have b - c = -3/2 SO
      (b - c)^3 = (-3/2)^3 = -27/8
      3(b - c) = 3(-3/2) = -9/2
      SO (b - c)^3 - 3(b - c) + 9/4 = -27/8 -3(-3/2) + 9/4
      = -27/8 + 9/2 + 9/4 = (-27 + 36 + 18)/4 = 27/8 = (3/2)^3
      (2 votes)
  • blobby green style avatar for user tomm0_
    Can you do this WITHOUT multiplying or dividing the equations first?
    I tried, but it isn't working.
    Also how do you know when you have modify them? If I came across a random equation, how would I know I need to multiple/divide it first?
    (5 votes)
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  • piceratops tree style avatar for user narichards12
    Is this also known as the combination method?
    (3 votes)
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  • blobby green style avatar for user Darlene Rodriguez
    Ask a question...can you cancel out the y instead of the x?
    (2 votes)
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  • purple pi purple style avatar for user souvik
    whats the difference between substitution and elimination methods in solving systems of equations
    (2 votes)
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  • leaf green style avatar for user K
    When you're multiplying/dividing to get the equations ready to cancel out a variable, do you have to multiply/divide both equations by something? Can't you just do it to one of the equations?
    (3 votes)
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  • winston default style avatar for user Nicole
    At around , how do I know when the fractions will work? Is it simply when everything is divisible by the fraction, or is there another way to figure it out?
    (1 vote)
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    • hopper happy style avatar for user Rachel
      It will always work, but first let me explain why, in this equation, it made sense to multiply both sides by fractions. By multiplying the 1st equation by 1/6 and the second equation by 1/5, the x's canceled out and it's easier to solve because ALL the variables ended up having 1 as a coefficient. I would not use fractions if it would create fractions in the equation, as fractions are a pain to work with. If the equation were, for instance,
      (1/6 )6x-4y=18 (1/6)
      (1/5 ) -5x-3y=20 (1/5)
      x-2/3y=3
      -x-3/5y=4
      I could multiply the first equation by 1/6 and go from a 6x to 1x and I could multiply the second equation by 1/5 and go from -5x to -1x. But then I would create fractions and it would be difficult to solve.
      (3 votes)

Video transcript

We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, that's not going to cancel it out. If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Negative 24 times 1/6 is negative 4. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things are equal. So we're doing the same principle that we saw when we first started looking to algebra, that you can maintain your equality as long as you add the same thing. On the left-hand side, we're going to add this. And on the right-hand side, we're going to add this. But this second equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on the left-hand side? Well, you have a positive x and a negative x. They cancel out. That was the whole point behind manipulating them in this way. And then you have negative y minus y, which is negative 2y. And then on the right-hand side, you have negative 4 minus 12, which is negative 16. And these are going to be equal to each other. Once again, we're adding the same thing to both sides. To solve for y, we can divide both sides by negative 2. And we are left with y is equal to positive 8. But we are not done yet. We want to go and substitute back into one of the equations. And we can substitute back into this one and to this one, or this one and this one. The solutions need to satisfy all of these essentially. This blue one is another way of expressing this blue equation. This green equation is another way of expressing this green equation. So I'll go for whichever one seems to be the simplest. And this one seems to be pretty simple right over here. So let's take x minus y-- we just solved that y would be positive 8-- is equal to negative 4. And now to solve for x, we just have to add 8 to both sides. And we are left with, on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8 is equal to positive 4. So you get x is equal to 4, y is equal to 8. And you can verify that it would work with either one of these equations. 6 times 4 is 24 minus 6 times 8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative 20, minus negative 40, if y is equal to 8, does, indeed, get you negative 60. So it works out for both of these. And we can try it out by inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good.