How do you solve the system y=x+7, y=(x+3)^2-8?

1 Answer
Phillip E.
Nov 15, 2015

There are two solutions x=1, y=8 and x=-6, y=1

Explanation:

Replace y in the second equation with the right hand side of the irst equation to give x+7=(x+3)^2-8. Expand the square on the right hand side x+7=x^2+6x+9-8. Subtract x+7 from both sides and tidy up x^2+5x-6=0. Factorise (x+6)(x-1)=0. This gives the two solutions x=1 and x=-6. Using the first equation, if x=1 then y=8 and if x=-6 then y=1.

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