Jamierox4ev3r
  • Jamierox4ev3r
Check my work someone, not sure if I'm going about this in the correct manner
Mathematics
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SOLVED
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chestercat
  • chestercat
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Jamierox4ev3r
  • Jamierox4ev3r
Give me a second to type out the problem, thank you
Jamierox4ev3r
  • Jamierox4ev3r
Simplify the rational expression: \(\Large\frac{10^{2}+29x+21}{x^{2}-25}\times\frac{x+5}{2x+3}\)
Jamierox4ev3r
  • Jamierox4ev3r
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)

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Jamierox4ev3r
  • Jamierox4ev3r
*factored both the numerator and denominator of the first part of the expression
Jamierox4ev3r
  • Jamierox4ev3r
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?
Jamierox4ev3r
  • Jamierox4ev3r
^This is where there are some clear holes in my recollection of simplifying. Someone clarify my thinking, thanks
jim_thompson5910
  • jim_thompson5910
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
Jamierox4ev3r
  • Jamierox4ev3r
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.
jim_thompson5910
  • jim_thompson5910
that's as simplified as it gets
Jamierox4ev3r
  • Jamierox4ev3r
Hmm alright. Thank you for your time :)
jim_thompson5910
  • jim_thompson5910
you're welcome

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