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Having trouble with the substitution method,

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I've attempted it several times but answers come out different each time. & I'm clueless to why.
What are the equations?
can you give us the equation for us to try it and see where the problem is?

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Other answers:

Its, |dw:1360705833711:dw|
My answers came out from 4 to somehow I got .14446 etc..
So I know Im doing it wrong somehow
Get the first equation equal to x. So x = 2y-5. Then sustitute that in.
(2y-5)^2 + y^2 - 25 = 0
4y^2 - 20y +25 - 25 = 0
5y^2 - 20y = 0
y=4 or y=0
I think alrightatmaths is right.
Thank you Kweku :)
Im lost
Would the 0 & 4 have a fraction beside?
Just make sure you have the variables on one side and the the whole numbers or fractions on the other side. {2y - x = 5 {2x + 2y - 25 = 0 In this case, subtract 2y on both sides and you should get -x = 5 - 2y. Then divide -1 due to the -x on both sides. You'll then have x = -5 + 2y. Now plug in the x = -5 + 2y into the second equation. This should help you solve for y. It'll end up being 2(-5 + 2y) + 2y - 25 = 0 Next it'll be substituted to -10 + 4y +2y -25 = 0. Put variables on one side and the whole numbers on the other side. The whole numbers are "negative," so you should now "add" 10 and 25 on both sides. You should get in the end, 6y = 35. Divide 6 on both sides. It should be fine if you get fractions or not. This is how I learned it. Not every set of system of equations are going to be easy. It may deal with whole numbers and/or fractions. Once you find one of the values such as y or x, then plug in the y or x value into the opposite equation that you didn't use to find that value. That's how it should be I believe.
Alrighty, I got it thanks.!

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