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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    @Astrophysics

  2. Astrophysics
    • one year ago
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    What method, substitution, elimination, matrices, etc?

  3. anonymous
    • one year ago
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    it doesnt tell me

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  4. Astrophysics
    • one year ago
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    You can solve for x by setting y = y so you get \[x^2+2x+7=x+7\]

  5. Astrophysics
    • one year ago
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    What do you get when you solve for x here?

  6. anonymous
    • one year ago
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    2x^3 i think

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

  8. Astrophysics
    • one year ago
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    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.

  9. Astrophysics
    • one year ago
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    \[x^2+x=0\] you can complete the square if you like

  10. Astrophysics
    • one year ago
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    Though you can just do \[x(x-1)=0\] now what are the roots?

  11. Astrophysics
    • one year ago
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    ^ 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.

  12. anonymous
    • one year ago
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    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

  13. Astrophysics
    • one year ago
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    @badsuns can you please stop giving out the answers.

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

  15. Astrophysics
    • one year ago
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    Thanks

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