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What method, substitution, elimination, matrices, etc?
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You can solve for x by setting y = y so you get \[x^2+2x+7=x+7\]
What do you get when you solve for x here?
2x^3 i think
Go to math way .com. It is also an app. Gives me almost every single math answer I can think of except word problems
No, you need to factor so you have \[x^2+2x+7=x+7 \implies x^2+2x+7-x-7=0 \implies x^2+x=0\] now solve for x.
\[x^2+x=0\] you can complete the square if you like
Though you can just do \[x(x-1)=0\] now what are the roots?
^ this will give your x values, once you find that just plug it in either of the original equations above to get the corresponding y value, it will be easier if you plug the x values in y = x+7, this will give you your (x,y) solutions where you should have two.
subtract x+7 on both sides of the equation
I think the answers are:
x = 1
x = 0
@badsuns can you please stop giving out the answers.
@Astrophysics just trying to help, but i'll stop. sorry about that. i'm not 100% sure on this answer anyway.