Algebra help: solving second degree equations?

Question

Hi guys so when solving with these types of equations there are times you can ONLY use the substitution method(depending on the equation) I think I made an error in this problem cause I never used the substitution method before to solve a system of equations can someone help please?

Here's the problem and my work that I did:

>x=2(y+1)^2-6 <---substitution can only be used since powers are different right?

x+y=3

>2(y+1)^2-6+y=3 <---plugged in for 'x'

>2(y^2+2y+1)-6+y=3

>2y^2+4y-4+y=3

>2y^2+5y=7

This is as far as I got I know I can take out a 'y' for the last step but aren't I supposed to make the constant on the right of the equation "0"??? Please help thanks!

Update:

@Vincent Yang: Oh okay I see now bring the 7 to the other side, then factor out. Thanks! When yahoo answers let's me I'll put yours as best answer

? · Accepted Answer

x = 2(y + 1)^2 - 6 
x + y = 3

x = 3 - y
3 - y = 2(y + 1)^2 - 6
3 - y = 2y^2 + 4y + 2 - 6
3 - y = 2y^2 + 4y - 4
0 = 2y^2 + 5y - 7
0 = (2y + 7) (y - 1)
y = -7/2 , 1

x + y = 3
x + (-7/2) = 3
x = 13/2 or 6 1/2
x + y = 3
x + 1 = 3
x = 2
Answers: 
(13/2, -7/2) and (2,1)

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Algebra help: solving second degree equations?

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