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8th grade (Eureka Math/EngageNY)

Unit 4: Lesson 4

Topic D: Systems of linear equations and their solutions

Systems of equations with elimination: 2x-y=14 & -6x+3y=-42

Sal solves the system of equations 2x - y = 14 and -6x + 3y = -42 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • leaf green style avatar for user Jin Hee Kim
    can there be a subtraction elimination?
    (20 votes)
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    • marcimus pink style avatar for user cara.ann.pomeroy
      Adding a negative and subtracting a positive are the same thing. Technically, with the existence of negative numbers, adding and subtracting are the same thing. You could,if you so desired, call this method "subtraction elimination" and just add a negative of the second equation; it would be the same process.
      (18 votes)
  • male robot johnny style avatar for user Fidel Rodriguez
    Can there be a subtraction elimination besides the ones there are?
    (1 vote)
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  • blobby green style avatar for user bobwburns
    He should not say that when x=1 or y=2 that they intersect in exactly one place. Just a little error. I may have missed something.
    (0 votes)
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  • aqualine ultimate style avatar for user inspirehaters
    i think i got the answer on accident
    it plugged them in and it seemed to work too
    y=-7
    x=3.5
    (1 vote)
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  • mr pants teal style avatar for user Isabella  D'Agostino
    What would the answer be if you had a problem such as 9x + y = -17 and -4x + y = -5? How would I continue the problem of eliminations if there is 5x + 2y = -22?
    (1 vote)
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    • leafers tree style avatar for user Neen
      Instead of adding the two equations, just subtract one equation from another to eliminate y >>> (9x+y)-(-4x+y) = -17-(-5) >>> 9x+y+4x-y = -12 >>> 13x = -12 >>> x = -12/13. Substitue the value of x in one of the equation to solve for the value of y (which when solved, y = -113/13) Hope this help :)
      (1 vote)
  • hopper cool style avatar for user Scott Sourile
    How would I go about using one of these 3 methods for a problem like this:

    Joe spent 201 dollars on shirts and pants. Shirts cost 27 each, and pants cost 22 each. Eight were bought in total. How many were bought of each kind?
    (1 vote)
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  • leafers seed style avatar for user qc19.crotenberger
    Sal, can there be a elimination where you subtract?
    (1 vote)
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  • leaf blue style avatar for user David
    why in the first exercise has to multiply x 3??.
    (1 vote)
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  • blobby blue style avatar for user Forrest Li
    At , is a dependent system means both equation are the same?
    (2 votes)
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    • male robot donald style avatar for user Admiral Betasin
      Partially. A dependent system is a system where both lines have the same slope.
      For example:
      x + y = 10
      2x + 2y = 20

      These are different equations, but the slopes are the same. This is because if you divide both sides by 2 in the second equation, you'll get x + y = 10.
      (0 votes)
  • mr pants teal style avatar for user marc-hayden-brewer
    I am faced with this problem...
    3x+5y =30
    5x+8y=49
    and I don't know how to start the problem
    (1 vote)
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    • starky sapling style avatar for user Hodorious
      If the two equations are part of a system you could solve that by elimination. One way would be to multiple the first equation by five and the second by negative three, so you'll have 15x and -15x cancelling each other out when you add the two equations. From there you'll just have to solve for y. Once you have the value of y, you can substitute it back to one of the original equations to solve for x.
      (1 vote)

Video transcript

Solve for x and y. And we have the system of equations right here. We have 2x minus y is equal to 14, and negative 6x plus 3y is equal to negative 42. So we could try to solve this by elimination, and let's see if we can eliminate our y variables first. We have a 3y here, and we have a negative y up here. And they won't eliminate. If you just add negative y plus 3y, that won't eliminate. But if we could turn this negative y into a negative 3y, then it would cancel out with the 3y. And the best way to turn a negative y into a negative 3y is to multiply this entire top equation by 3. So let's do that. Let me get some space over to the left Let's multiply this entire top equation by 3. So I'm going to multiply it by 3. So I multiply 2x by 3. I get 6x. I multiply negative y by 3. I get negative 3y. And then I multiply 14 by 3. 3 times 14 is 42, right? 3 times 10 is 30, plus 12, it's 42. And then we can add both of these equations. Something interesting should already maybe be showing up on your radar. But let's add both of these equations. Let's add the left-hand side. Negative 6x plus 6x? Well, those cancel out. We get 0. Then we have 3y minus 3y. Those cancel out. We get another 0. And then finally, you get negative 42 plus 42. Well, that's a 0. So we end up with just 0 is equal to 0, which is clearly true, but it's not putting any constraints on the x or y. And that's because whenever you have a situation like this, where you just get something that's obviously true-- 0 equals 0, or 1 equals 1, or 5 equals 5-- what we're dealing in this situation is where both of our constraints, both of our equations, are actually the same equation. So this right here is a dependent system. And you see it right over here. If you take that first equation, you multiply it by 3, you got 6x minus 3y is equal to 42. If we then multiplied it by negative 1, you would get the exact same equation as the second equation. You would get negative 6x plus 3y is equal to negative 42. Or another way to think about it, if you want to go from the first equation to the second equation, you just multiply both sides of the equation times negative 3. So both of these constraints are actually the same constraints. They're just kind of a scaled-up multiple of each other. So if you were to graph them-- and I might as well graph them for you right here. This first equation right here, let me do it over here. It's 2x minus y is equal to 14. You could subtract 2x from both sides, and you would get-- let me just subtract it. Subtract 2x, subtract 2x. On the left-hand side, you're left with just negative y. On the right-hand side, you have negative 2x plus 14. Multiply both sides by negative 1, and you get y is equal to 2x minus 14. So this first equation over here, if I were to draw my axes-- so that is my y-axis, and then that is my x-axis. This graph right over here, its y-intercept is negative 14. So this is 0, negative 14. And it has a slope of 2. So it's going to look something like this. It's going to look something like that. And then this second equation, if you were to graph it, because this is a dependent system, it's the exact same line. If you were to put this in slope-intercept form and graph it, you would get the exact same thing. It would go right on top of it. So there's actually an infinite number of solutions. These two lines are the same line. So they intersect everywhere on each of the respective lines. They're the same line. And when you get something like this, 0 equals 0 or 1 equals 1, that's the telltale sign that you're dealing with a dependent system. If you got something like 0 equals 1, then you would have no solution. So this one right here would be an inconsistent system. And this would be the situation. So let me make it very clear. This is the situation. We're dealing with the same lines. The lines are the same lines. This is a situation where you have parallel lines. And so they never intersect. And then, obviously, you have the easiest situation. I think this is one most of us are familiar with, where you have something like x is equal to 1, or y is equal to 2, or anything like this. It doesn't have to be 1 or 2. And then this is clearly a situation where you have an independent system, where you have two different lines that intersect in exactly one place. Anyway, this-- there's an infinite number of solutions. Any x and y that satisfied the first equation will also satisfy the second equation, because they're fundamentally the same constraint.