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.

# 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.

## Want to join the conversation?

• 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?
• You are trying to keep the two sides of the equations equal. The equations themselves are not, and do not need to be equal.

if 2 = 2
& 5 = 5 even though these are not equal you can add them
->7 = 7 and since the two sides were equal before they will still be equal after you add them.
• 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..
• 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)
• 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
• 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.
• 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"
• 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
• 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?
• You will know you won't have to modify them if one variable will cancel out during substitution. For example, 5x+3y=10 and -5x+-8y=-14 won't have to be modified because the 5x and -5x will cancel.
(1 vote)
• Is this also known as the combination method?
• Ask a question...can you cancel out the y instead of the x?
• Yes you can. That would mean that you solve for x first and then substitute the value to get y. Same method.
• whats the difference between substitution and elimination methods in solving systems of equations
• Substitution is when you isolate a variable and substitute it into another equation. Elimination is when you add or subtract an equation to another.
• 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?
• You can do anything to just one equation, as long as you do it to both sides of it.
(1 vote)
• 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)
• 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.