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# Solving systems of equations by elimination (old)

An old video where Sal introduces the elimination method for systems of linear equations. Created by Sal Khan.

## Want to join the conversation?

• for the first problem... the 4y= -8........ where did the -8 came from?
• After finding the value of x= ⁷⁄₂, he had:
3x + 4y = ⁵⁄₂
Putting the x= ⁷⁄₂ in for x we get:
(3)(⁷⁄₂) + 4y = ⁵⁄₂
Multiplying the 3 and the ⁷⁄₂ gives:
²¹⁄₂ + 4y = ⁵⁄₂
Subtracting ²¹⁄₂ from both sides gives:
4y = ⁵⁄₂ - ²¹⁄₂
Combining like fractions:
4y = ⁵⁻²¹⁄₂
Since 5-21=-16, we get:
4y = -16/2
Since -16/2 = -8 we get
4y = -8
Dividing by 4 gives us:
y = -2
• Hey Sal, how can solve a system of equation with the elimination IF you can't cancel a variable?
EX: 5x+3y=12 and 4x-5y=17
-Thanks!
• One way you can do that is by multiplying the top equation by 5 and multiplying the bottom equation by 3 because then, you could easily cancel out the 15 (top equation) and the -15 (bottom equation) and solve the rest of the equation accordingly.
>>>

5x + 3y = 12
4x - 5y = 17

5x + 15y = 12
4x - 15y = 17

5x = 12
4x = 17

9x = 29

x = ~3.22

(Then solve for y)
Hope this helps!
(1 vote)
• -4x-2y=-12
4x+8y=-24
• first you have to subtract from both sides . Then you have to divide the whole equation by whatever your number is. Then you would eventually get down to a new dividing processes. Next you would divide and find your answer. After you are done with your steps then you would have to go back into your original equation and plug it in for your letter Y.
• Why are there letters in math it is bummy and shouldnt exist
• Well technically they're not letters-they represent unknown variables, so technically, you can use any other character in PLACE of the letter. We just chose letters to represent the unknown.
• Im kind of stuck so if i had an equation like... 4b+3v=29
6b+3v=39
How would i solve this problem??
• Both equations have the term "3v". But, the signs are the same. If you make one have "-3v", then you can eliminate the "v" variable and solve for "b". Here's how to do it:
1) Multiply one of the 2 equations by -1. For example:
-1 (4b+3v) = -1(29)
-4b - 3v = -29
2) Add the 2 equations to eliminate "v"
6b + 3v - 4b - 3v = 39 - 29
2b = 20
3) Solve for "b" by dividing by 2: b = 10
4) Then, use the value of "b" to find the value of "v" by substituting back into one of the equations.

Hope this helps.
• What do you do if your system has no same values ie
4x+y-2z=-6
2x-3y+3z=9
x-2y=0
• How would you do something like. -6x + 3y = -18 and -3x + 4y = 6 ?
• There are a few ways to solve this, but I'll tell you how I did it. Since I find graphing my equations easier, I decided to put both these equations in y=mx+b form. For -6x+3y=-18, solve for y by adding 6x to both sides, and you get 3y = 6x + 18. Divide all by 3 and your first graphable equation is y=2x+6. Now you have to convert the other equation,-3x+4y=6. Add 3x to both sides and end up with 4y=3x+6. Divide out by 4, and your second equation should equal y=3/4x+1.5. Once you graph it, the lines should intersect at about the point (-2,2) or (-2,2.5). Probably not the method you're looking for, but I hope it still helps anyway :)
• For the last question you would simplify subtract the top equation from the bottom equation because you can learn the rule SSS. Same Signs Subtract. Hope this helps for anyone. btw i am in grade 8 :)
• I know three easy steps to solve these type of equations by elimination method: