My Math Forum Dividing two fractions with equations in the numerators

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 May 22nd, 2015, 04:41 PM #1 Member   Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0 Dividing two fractions with equations in the numerators $\displaystyle \frac{4x+8}{15} \div \frac{3x+6}{6}$ I changed the equation by turning it into multiplication and changing the second fraction to its reciprocal: $\displaystyle \frac{4x+8}{15} \times \frac{6}{3x+6}$ I divided the second fraction by $\displaystyle \frac {3}{3}$: $\displaystyle \frac{4x+8}{15} \times \frac{2}{x+2}$ but multiplying those together just confused me more, because I got a numerator divisible by 4 and a denominator divisible by 15, but, as far as I can see, I can't do anything with either of them: $\displaystyle \frac{8x+16}{15x+30}$ So I'm stuck. The answer is $\displaystyle \frac{8}{15}$, and I don't see how to get rid of the x's or the 16 and 30. Let me know if you find anything strange; I might have made a typo.
 May 22nd, 2015, 05:25 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,467 Thanks: 2038 $\displaystyle \frac{4x + 8}{15} \times \frac{2}{x+2} = \frac{4(\cancel{x + 2})}{15} \times \frac{2}{\cancel{x + 2}} = \frac{8}{15}$ Thanks from Rexan
May 22nd, 2015, 06:19 PM   #3
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 Originally Posted by skipjack $\displaystyle \frac{4x + 8}{15} \times \frac{2}{x+2} = \frac{4(\cancel{x + 2})}{15} \times \frac{2}{\cancel{x + 2}} = \frac{8}{15}$
Ah. Thanks for pointing that out. To catch that sort of thing in future problems, I'll just have to be super observant?

 May 22nd, 2015, 08:27 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 It's usually a good idea to factor the numerator and denominator of your fraction. That will make it clear as to whether there are any common factors. Thanks from Rexan
May 22nd, 2015, 08:59 PM   #5
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 Originally Posted by Azzajazz It's usually a good idea to factor the numerator and denominator of your fraction. That will make it clear as to whether there are any common factors.
Okay, that makes a lot sense. Thank you!

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