If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

8th grade (Eureka Math/EngageNY)

Unit 4: Lesson 4

Topic D: Systems of linear equations and their solutions

Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120

Learn to solve the system of equations y = -1/4x + 100 and y = -1/4x + 120 using substitution.  Created by Sal Khan and Monterey Institute for Technology and Education.

Want to join the conversation?

  • blobby green style avatar for user Deandre Shielders
    Wht about a problem like this
    -5x+y=-3
    3x-8y=24
    (8 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Eason The Bestest
      Let -5x+y=-3 be the first equation, and 3x-8y=24 be the second. Let's manipulate the first equation by adding 5x to both side; resulting in y=-3+5x, or 5x-3. Now let's plug in 5x-3 for wherever y has been in the second formula. 3x-8(5x-3)=24. This is now pretty straightforward to solve for x. We distribute the 8 into (5x-3) and get 3x-40x-24=24. -48=27x. x=-48/27, x=-16/9.

      Now let's plug in the x for either one of the equations to get y. 3(-16/9)+8y=24. -16/3+8y=24. 8y=24+16/3. 24y=72+16. 24y=88. y=88/24, y=11/3.

      Hope it helped!
      (2 votes)
  • blobby green style avatar for user Regina Saffo
    solve for y in terms of x
    5x +380y= -2280
    (4 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Chuck Towle
      Regina,
      Solve for y in terms of x just means "isolate y on one side of the equation".

      You do the same thing to both sides of the equation to keep it equal while eliminating everything but the y on the left.

      If you first subtract 5x from both sides, the left will have 5x-5x which becomes 0 and disappears leaving you with only 380y on the left.

      If you then multiply each side by the reciprocal of 380 (which is 1/380, the 380/380 on the left becomes 1 and disappears leaving you with only y on the left.

      Make sure as you do this, you do the same thing to the right side.

      When the y becomes isolated on the left, you will have solved for y in terms of x.

      I hope that helps make it click for you.
      (4 votes)
  • blobby green style avatar for user Talia Wadsworth
    How do you know what you're supposed to multiply the equations by?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Matthew
    I am still a little bit confused about why the substitution method actually works. Can someone please explain it to me more?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Gabby Aase-Remedios
    How do you solve with x/3-y/2=1 and y=2x-2
    (3 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Chuck Towle
      Gabby,
      x/3-y/2=1
      y=2x-2
      From the second equation you know that y and 2x-2 are the same so you can substitute (2x-2) for the y in the first equation
      x/3-y/2=1 Substiture (2x-2) for the y
      x/3 - (2x-2)/2 = 1
      Now distribute the -1/2
      x/3 - 2x/2 + 2/2 = 1 The 2/2s both become 1
      x/3 - 1x +1 = 1 Subtract 1 from both sides
      x/3 - 1x = 0
      Now, you might just see that x=0 is the solution for x,
      Or to combine the x terms, you must find a common denominator.
      x/3 - (3/3)x = 0 now 1/3 - 3/3 = -2/3 so
      (-2/3)x = 0 Now multiply both sides by the reciprocal of -2/3 which is -3/2
      x=0

      Now find y by substituting 0 for x in either equation.
      y= 2(0) -2
      y= 0-2
      y= -2

      Now double check by substituting 0 for x and -2 for y in the other equation
      0/3 - (-2/2) = 1
      0-(-1) = 1
      1=1
      This is true so the answers double check.
      x=0 y=-2 or (0,-2)

      I hope that helps make it click for you.
      (3 votes)
  • winston default style avatar for user Naomi Tarumoto
    So, what if the X was solved already for us and the Y would be the thing we would have to solve. Would this apply the same way it does for solving X? Solving Y?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Zachary Hamilton
      Yes, solving Y is the same thing as solving X. Lets say that you have x already solved lets say x=3. Now you have to solve for Y. It is the exact same thing. Lets plug what we know into an equation. Lets say that 3+4y=15. Do the same thing you would to find X. If you do your same steps, you will get that Y=4 even doing the same thing you use for X.
      (1 vote)
  • blobby green style avatar for user William Korn
    How do you solve this problem 4x+3y=2 and 2x+y=6?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user robertschmidt54
      Line them up like this:
      4x+3y=2
      2x+y=6
      Now ask yourself can i easily solve for x or y, or can I somehow get rid of an x or y?
      I see an easy way to solve for y in the second equation:

      2x+y=6 therefore y= -2x+6

      now we have a function for y in terms of x we can then place it in the top equation and solve for x

      4x+3(-2x+6)=2 => 4x-6x+18=2 => -2x=-16 therefore x=8

      we can then take this value for x and solve for y using any of the two equations but the second one is more convenient

      y= -2x+6 = -2(8)+6=-16+6= -10 therefore y= -10

      we can check these results

      2(8)-10=6 => 16-6=10 => 10=10 check
      4(8)+3(-10)=2 => 32-30=2 => 2=2 check
      (6 votes)
  • blobby green style avatar for user Melissa White
    What is x-y=12
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user jnavhoopz
    So how would I solve an equation like
    y= 6x - 14
    y= -8x
    ?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user henrysp3
      Since the two equations are equal to the same thing (y) you can set the two equations equal to each other (6x - 14 = -8x). By bringing the variables over to the right side and 14 to the left we get 14x = 14. Simply divide both sides by 14 to get x=1.
      (4 votes)
  • aqualine ultimate style avatar for user PreXash18stiffler
    im confused of ho to do all these
    (1 vote)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Barrett Southworth
      A variable can be viewed as a container, when it comes to say two different equalities involving x and y. Substitution means finding what x or y is equal to in one equation and replacing that x or y in the other with the other side of the equal sign in the first.
      Lets say the first equation is
      2x + 3y = 12
      3y = 12 - 2x
      y = 6 - 2x/3
      There is a second equation of oh lets say
      x - y = 3, the y can be replace with (6 - 2x/3)
      x - (6 - 2x/3) = 3, distributing out negative sign
      x + 2x/3 - 6 = 3
      (5/3)x = 9
      x = 27/5
      (4 votes)

Video transcript

We're given a system of equations here, and we're told to solve for x and y. Now, the easiest thing to do here, since in both equations they're explicitly solved for y, is say, well, if y is equal to that, and y also has to equal this second equation, then why don't we just set them equal to each other? Or another way to think about it is, if y is equal to this whole thing right over here-- that's what that first equation is telling us-- and if we have to find an x and a y that satisfy both of these equations, if y is equal to that, why can't I just substitute that right here for y? And if we do that, the left-hand side of this bottom equation becomes negative 1/4x plus 100. And then that is going to be equal to this right-hand side-- and I'll do it in the same color-- is equal to negative 1/4x plus 120. Now, the first thing we might want to do is maybe get all of our x terms onto the left- or the right-hand side of the equation. And if we wanted to get rid of these x terms from the right-hand side, get them on the left-hand side, the best thing to do is to add 1/4x to both sides of this equation. So let me do that. So we're going to add 1/4x here, add 1/4x here, and you might already be sensing that something shady is going on. So let's do it. So negative 1/4x plus 1/4x. They cancel out. You get 0x. So the left side of the equation is just 100. And then the right side of the equation, same thing. Negative 1/4x plus 1/4x. They cancel out. No x's. And you're just left with is equal to 120. Which we know is definitely not the case. 100 is not equal to 120. We got this nonsensical equation here, that 100 equals 120. So this type of system has no solution. You know it has no solution because in order for it to have any solution, these two numbers would have to be equal to each other, and they are not equal to each other. And if you look at the original equations, it might jump out at you why they have no solutions. Both of these lines, or both of these equations, if you view them as lines, have the exact same slope. But they have different y-intercepts. So if I just were to do a really quick graph here. That's my y-axis, that is my x-axis, so it's y and x. This first graph over here, its y-intercept is 100. Let me do it a little bit lower. Its y-intercept-- let's say that that is 100, so it intersects right there. And there's a slope of negative 1/4. So maybe it looks something like this. That's that first line. This second line-- I'll do it in pink right here-- y is equal to negative 1/4x plus 120, its y-intercept might be right here at 120. But it has the same slope, negative 1/4, so its slope, the line would look something like this. So you see that there are no x and y points that satisfy both of these equations. Another way to think about it. If y-- you take an x. This first equation says, OK, you take your x, multiply it by negative 1/4, and add 100, and that's going to give you y. Now, here we say, well, you take that same x, and you multiply it by negative 1/4 and add 120, and that has to be equal to y. Well, the only way that that would ever be true is if 100 and 120 were the same number, and they're not the same number. So you're never going to have a solution of this system. These two lines are never going to intersect, and that's because they have the exact same slope.