# math

(Systems) how do I solve 5x+3y=2 and x+2y=6 using substition

This is what I did

X+2y=6 eq.1
5x+3y=2 eq.2
X=-3y/5

Sub(x=-3y/5) into eq.1
-3y/5 + 2y=6
7/5y=6
y=30/7

Sub(y=30/7) into eq.2
5x+3(30/7)=3
5x+90/7=3
5x=-69/7
x= -68/35

The answer I got was (-68/35, 30/7) but I know it's wrong because there were no fractions in the possible answers.

I know how to solve it with elimination but I can't figure out how to do it with substitution.

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1. Why not take the easier equation?
x+2y=6 ---> x = 6-2y

now plug that into the other,
5x+3y=2
5(6-2y) + 3y = 2
30 - 10y + 3y = 2
-7y = -28
y = 4
then x = 6 -2(4) = -2

see how easy it is ?

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2. Thank u!

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3. Your mistake happened early. You said

5x+3y=2 eq.2
X=-3y/5

but really

x = (2-3y)/5

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