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Jamierox4ev3r

  • one year ago

Check my work someone, not sure if I'm going about this in the correct manner

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  1. Jamierox4ev3r
    • one year ago
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    Give me a second to type out the problem, thank you

  2. Jamierox4ev3r
    • one year ago
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    Simplify the rational expression: \(\Large\frac{10^{2}+29x+21}{x^{2}-25}\times\frac{x+5}{2x+3}\)

  3. Jamierox4ev3r
    • one year ago
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    So far, I've factored the top portion. So \(10x^{2}+29x+21\) should turn into (5x+7)(2x+3) while \(x^{2}-25\) should be (x-5)(x+5)

  4. Jamierox4ev3r
    • one year ago
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    *factored both the numerator and denominator of the first part of the expression

  5. Jamierox4ev3r
    • one year ago
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    So I tried cross-multiplying, but that brought me more complications...however, I noticed a few things \(\huge\frac{(5x-7)(2x+3)}{(x-5)(x+5)}\times\frac{x+5}{2x+3}\) \(\huge\frac{(5x-7)\color{red}{(2x+3)}}{(x-5)\color{red}{(x+5)}}\times\frac{\color{red}{x+5}}{\color{red}{2x+3}}\) Can the things in red cancel out? And would that lead me with a final answer?

  6. Jamierox4ev3r
    • one year ago
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    ^This is where there are some clear holes in my recollection of simplifying. Someone clarify my thinking, thanks

  7. jim_thompson5910
    • one year ago
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    you are correct. The `2x+3` terms will divide to 1 and effectively go away since 1*(any number) = same number the same applies to the `x+5` terms

  8. Jamierox4ev3r
    • one year ago
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    so \(\huge\frac{5x+7}{x-5}\) would be final answer? I mean, I'm not sure if that could be simplified any further, but I don't think so.

  9. jim_thompson5910
    • one year ago
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    that's as simplified as it gets

  10. Jamierox4ev3r
    • one year ago
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    Hmm alright. Thank you for your time :)

  11. jim_thompson5910
    • one year ago
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    you're welcome

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