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anonymous

  • 5 years ago

One number is 44 more than twice another. Their product is 44 more than twice their sum. Find the numbers x and y. (Assume x < y.)

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  1. anonymous
    • 5 years ago
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  2. amistre64
    • 5 years ago
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    m = 44+2n mn = 44+2(m+n)

  3. anonymous
    • 5 years ago
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    y=44+2x y=2(x+y)+44 Solve system using substitution or elimination

  4. amistre64
    • 5 years ago
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    mn = 44 +2m+2n ; yet 2n = m-44 mn = 44 +2m +2m -88 mn = -44 +4m 44 = mn +4m 44 = m(n+4) m = 44/(n+4) ; n!= -4

  5. amistre64
    • 5 years ago
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    m = 44+2n = 44/(n+4) (44+2n)(n+4) = 44 44n +176 +2n^2 +8n = 44 2n^2 +52n +176-44 = 0 2n^2 +52n +132 = 0

  6. amistre64
    • 5 years ago
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    2(n^2 +26n + 66) = 0 n^2 +26 = -66 (n+13) = sqrt(-66 + 169) n = -13 +- sqrt(103)

  7. amistre64
    • 5 years ago
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    htats prolly wrong lol

  8. anonymous
    • 5 years ago
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    it is. i know there's no square root for any of the problems. this one is killing me! haha

  9. anonymous
    • 5 years ago
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    one of the numbers is 0.

  10. amistre64
    • 5 years ago
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    m = 44+2n mn = 44+2(m+n) these are good equations right?

  11. amistre64
    • 5 years ago
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    if one of the numbers is 0; then their product is 0

  12. anonymous
    • 5 years ago
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    y=44+2x y=2(x+y)+44 The solution to this system is x=-22 y=0

  13. amistre64
    • 5 years ago
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    how is y a product in that?

  14. amistre64
    • 5 years ago
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    -22 is a good solution, I just cant see how you interpret the information..

  15. anonymous
    • 5 years ago
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    we'll skip that one. lol

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