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anonymous

  • one year ago

Can someone check my answer? Algebra II

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  1. anonymous
    • one year ago
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    \[2x+5/x^2-3x-10 \left( + \right) x+1/x+2\]

  2. anonymous
    • one year ago
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    I got the answer ...\[x(x-2)/(x-5)(x+2)\]

  3. anonymous
    • one year ago
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    @Hero

  4. anonymous
    • one year ago
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    @Michele_Laino @nincompoop

  5. anonymous
    • one year ago
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    @phi

  6. anonymous
    • one year ago
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    @texaschic101

  7. phi
    • one year ago
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    is this \[ \frac{2x+5}{x^2-3x-10} + \frac{ x+1}{x+2} \]?

  8. anonymous
    • one year ago
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    yes @phi

  9. phi
    • one year ago
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    first factor the bottom of the first fraction (to get a better idea of what is going on) \[ \frac{2x+5}{x^2-3x-10} + \frac{ x+1}{x+2} \\ \frac{2x+5}{(x+2)(x-5)} + \frac{ x+1}{x+2} \]

  10. anonymous
    • one year ago
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    That is what I did

  11. phi
    • one year ago
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    to get a common denominator, multiply the second fraction by (x-5) (top and bottom) \[ \frac{2x+5}{(x+2)(x-5)} + \frac{ (x+1)}{(x+2)} \frac{(x-5)}{(x-5)} \] now we can add the tops. First, multiply out the top of the second fraction: \[ \frac{2x+5}{(x+2)(x-5)} + \frac{ x^2-4x-5} {(x+2)(x-5)}\]

  12. phi
    • one year ago
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    now add the tops and put the sum over the common denominator \[\frac{ x^2-2x} {(x+2)(x-5)}\] it looks like we can factor out an x \[ \frac{ x(x-2)} {(x+2)(x-5)}\] yes, it looks like the same thing as what you posted.

  13. anonymous
    • one year ago
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    awesome! thanks for helping me out so quickly! @phi

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