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raebaby420

Please help.. What are the solutions to the following system of equations? y = x2 + 12x + 30 8x − y = 10 (−4, −2) and (2, 5) (−2, −4) and (2, 5) (−2, −4) and (5, 2) No Real Solutions

  • 7 months ago
  • 7 months ago

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  1. raebaby420
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    @Nurali

    • 7 months ago
  2. uri
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    @DebbieG Help her ^_^ Uri is proud of you.

    • 7 months ago
  3. DebbieG
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    What methods have you learned for solving a system? Substitution, elimination?

    • 7 months ago
  4. raebaby420
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    Elimination I think

    • 7 months ago
  5. raebaby420
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    And some substitution but I cant remember how to do that method

    • 7 months ago
  6. DebbieG
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    I would probably opt for substitution here, since you have a quadratic (a parabola) and a linear equation (a line). But either would work. If you solve the 2nd equation for y, then you have two equations that both give y as a function of x, right? E.g., you have y={some stuff} and y={some different stuff} so since y= both of those things, you can set them equal to each other.

    • 7 months ago
  7. DebbieG
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    When you do that, you'll have a single, quadratic equation to solve.

    • 7 months ago
  8. raebaby420
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    ..... that makes no sense

    • 7 months ago
  9. DebbieG
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    What part didn't make sense? Here are your two equations: \[y=x^2+12x+30\]\[8x-y=10\]Can you solve the 2nd one for y?

    • 7 months ago
  10. raebaby420
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    Idk how.. I only know how to solve the ones with Y on the other side of the equal sign

    • 7 months ago
  11. DebbieG
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    You just need to rearrange 8x-y=10 so that the y is by itself. Try adding y to both sides: 8x-y+y=10+y What do you get when you simplify that? Then subtract 10 from both sides.

    • 7 months ago
  12. raebaby420
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    IM SO FREAKING LOST

    • 7 months ago
  13. raebaby420
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    THANKS GUYS. I don't need any more help

    • 7 months ago
  14. DebbieG
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    @jessssloukaa the idea is not to do the problem FOR her.

    • 7 months ago
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