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sammy12 Group Title

If xy = 6 and x^2 + y^2 = 16, then what is the value of (x + y)?

  • 2 years ago
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  1. ChrisV Group Title
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    X+y=5

    • 2 years ago
  2. sammy12 Group Title
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    please can you explain

    • 2 years ago
  3. ChrisV Group Title
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    im trying to remember how to do this one sec

    • 2 years ago
  4. lazypig Group Title
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    \[x ^{2}+y ^{2}+2xy =16+6\times2\] \[\left( x +y \right)^{2}=28\] \[x +y =\pm \sqrt{28}\]

    • 2 years ago
  5. ChrisV Group Title
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    dunno how you came up with that but no way that works

    • 2 years ago
  6. sammy12 Group Title
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    Thanks

    • 2 years ago
  7. ChrisV Group Title
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    its obvious the values of x and y are 2 and 3 but i cannot for the life of me remember how to explain it

    • 2 years ago
  8. ChrisV Group Title
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    so x+y=5

    • 2 years ago
  9. myininaya Group Title
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    \[(x+y)^2=x^2+2xy+y^2=x^2+y^2+2(xy)=16+2(6)=16+12=28\] so we have \[(x+y)^2=28 => x+y=\pm \sqrt{28}\]

    • 2 years ago
  10. myininaya Group Title
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    so i agree with lazy :)

    • 2 years ago
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