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Mathematics 7 Online
OpenStudy (anonymous):

What is the solution set of y=x2+2x+7 and y=x+7?

OpenStudy (anonymous):

@Astrophysics

OpenStudy (astrophysics):

What method, substitution, elimination, matrices, etc?

OpenStudy (anonymous):

it doesnt tell me

OpenStudy (astrophysics):

You can solve for x by setting y = y so you get \[x^2+2x+7=x+7\]

OpenStudy (astrophysics):

What do you get when you solve for x here?

OpenStudy (anonymous):

2x^3 i think

OpenStudy (anonymous):

Go to math way .com. It is also an app. Gives me almost every single math answer I can think of except word problems

OpenStudy (astrophysics):

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.

OpenStudy (astrophysics):

\[x^2+x=0\] you can complete the square if you like

OpenStudy (astrophysics):

Though you can just do \[x(x-1)=0\] now what are the roots?

OpenStudy (astrophysics):

^ 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.

OpenStudy (anonymous):

x^2+2x+7=x+7 subtract x+7 on both sides of the equation x^2+x=0 I think the answers are: x = 1 x = 0

OpenStudy (astrophysics):

@badsuns can you please stop giving out the answers.

OpenStudy (anonymous):

@Astrophysics just trying to help, but i'll stop. sorry about that. i'm not 100% sure on this answer anyway.

OpenStudy (astrophysics):

Thanks

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