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by **Kprime911** on Sun Apr 22, 2012 8:36 pm

I am having trouble with this multiplying rational expression problem. answers is also suppose to be reduced to lowest terms.

3x-6/6x^2 * 9x/4x-8

thank you any help is appreciated.

**Sponsor**

by **stapel_eliz** on Mon Apr 23, 2012 11:19 am

Kprime911 wrote:3x-6/6x^2 * 9x/4x-8

What you have written means the following:

. . . . . $3x\, -\, \frac{6}{6x^2}\, \times\, \frac{9x}{4x}\, -\, 8$

Is this what you meant? Or were the grouping symbols omitted in [(3x - 6)/(6x^2)] * [(9x)/(4x - 8)], which is the following:

. . . . . $\left(\frac{3x\, -\, 6}{6x^2}\right)\, \times\, \left(\frac{9x}{4x\, -\, 8}\right)$

Or is the expression something else? Thank you! :wink:

by **Kprime911** on Mon Apr 23, 2012 4:48 pm

oh sorry it was the second one. [(3x - 6)/(6x^2)] * [(9x)/(4x - 8)].

by **little_dragon** on Mon Apr 23, 2012 10:30 pm

. . . . . $\left(\frac{3x\, -\, 6}{6x^2}\right)\, \times\, \left(\frac{9x}{4x\, -\, 8}\right)$

U can see how to do these here: http://www.purplemath.com/modules/rtnlmult.htm
start with factoring: $\left(\frac{3(x\, -\, 2)}{6x^2}\right)\, \times\, \left(\frac{9x}{4(x\, -\, 2)}\right)$
cancel the 3 w/ the 6 & both x-2s
times whats left
:wave:

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