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geerky42

  • 3 years ago

If \(\dfrac{x-8}{x+3}+\dfrac{x+3}{x-8}\) is equal to the mixed number \(A \dfrac{B}{(x+3)(x-8)}\), then B = ...

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  1. geerky42
    • 3 years ago
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    A. 64 B. 8 C. 121 D. 24 E. 9

  2. geerky42
    • 3 years ago
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    So far, I have B + A(x+3)(x-8) = (x+3)²+(x-8)² I don't know what to do next...

  3. geerky42
    • 3 years ago
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    Apparently, I need second equation somehow.

  4. ZeHanz
    • 3 years ago
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    OK, so we add the fractions:\[\frac{ x-8 }{ x+3 }+\frac{ x+3 }{ x-8 }=\frac{ (x-8)^2+(x+3)^2 }{ (x+3)(x-8) }\]\[=\frac{ x^2-16x+64+x^2+6x+9 }{ (x+3)(x-8) }=\frac{ 2x^2-10x+73 }{ x^2-5x-24 }=\] From this it is clear that A=2. Now we have to get B!

  5. ZeHanz
    • 3 years ago
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    The last fraction is:\[\frac{ 2x^2-10x+73 }{ x^2-5x-24 }=\frac{ 2x^2-10x-48+121 }{ x^2-5x-24 }=\frac{ 2(x^2-5x-24)+121 }{ x^2-5x-24 }\]\[=2\frac{ 121 }{ x^2-5x-24 }=2\frac{ 121 }{ (x+3)(x-8) }\]So B = 121.

  6. geerky42
    • 3 years ago
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    @ZeHanz

  7. ZeHanz
    • 3 years ago
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    @geerky42: Thx!

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