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britbrat4290

Solve the system by the substitution method. xy = 12 x2 + y2 = 40

  • 8 months ago
  • 8 months ago

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  1. live10000000
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    Why can't you do your own work

    • 8 months ago
  2. britbrat4290
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    y dont u pop off

    • 8 months ago
  3. live10000000
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    no thanks

    • 8 months ago
  4. britbrat4290
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    yea get off my question if u just wanna start trouble toodels

    • 8 months ago
  5. ilfy214
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    Thats hilarious sorry! :) But anyways you can use either equation: \[xy = 12 \]\[x= \frac{ 12 }{ y }\] OR \[x^2 + y^2 = 40 \]\[x + y = \sqrt{40}\]\[x = \sqrt{40} - y\] **The y and the x can be switched....

    • 8 months ago
  6. ilfy214
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    All you need to do is plug in.

    • 8 months ago
  7. britbrat4290
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    haha i know .. ok then y does it give me 4 quandrants to pick from ?

    • 8 months ago
  8. britbrat4290
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    example {( 2, 6), ( 6, 2), ( 2, -6), ( 6, -2)}

    • 8 months ago
  9. britbrat4290
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    @ilfy214

    • 8 months ago
  10. ilfy214
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    From what I solved I got these solutions. 6, 2, -2

    • 8 months ago
  11. britbrat4290
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    im just confused on plugging them in ..

    • 8 months ago
  12. ilfy214
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    Doesn't it depend on how you solved your equation? Also, when I plugged in -2 and 6 for the first problem, I got -12 instead of 12.

    • 8 months ago
  13. britbrat4290
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    im going w/ b

    • 8 months ago
  14. ilfy214
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    I was thinking for along the lines of C?

    • 8 months ago
  15. britbrat4290
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    i just dont understand how i allways mess that up

    • 8 months ago
  16. ilfy214
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    Because in B, if you try using ( 2, -6) and ( -2, 6) you already have a problem for the equation below:\[xy = 12\]\[(2)(-6) = -12\]\[(-2)(6)= -12\]

    • 8 months ago
  17. ilfy214
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    Its okay! Just choose the options that give you BOTH positive answers

    • 8 months ago
  18. britbrat4290
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    thank u i appreciate it !! :)

    • 8 months ago
  19. ilfy214
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    You're welcome! :D

    • 8 months ago
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