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 September 10th, 2014, 05:34 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,954 Thanks: 132 Math Focus: Trigonometry Algebraic Fraction Addition 5. The simple form of $\displaystyle \frac{4xy}{5}+\frac{15x}{2y^2} is ....$ A. $\displaystyle \frac{6x}{y^2}$ B. $\displaystyle \frac{60xy}{10y^2}$ C. $\displaystyle \frac{6x^2}{y^2}$ D. $\displaystyle \frac{6x^2}{y}$ I did it like this: $\displaystyle \frac{4xy}{5}+\frac{15x}{2y^2}$ $\displaystyle =\frac{2y^2(4xy)}{10y^2}+\frac{5(15x)}{10y^2}$ $\displaystyle =\frac{8xy^3+75x}{10y^2}$ Got stuck there. Can it be simplified further?
September 10th, 2014, 06:15 PM   #2
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 Originally Posted by Monox D. I-Fly 5. The simple form of $\displaystyle \frac{4xy}{5}+\frac{15x}{2y^2} is ....$ A. $\displaystyle \frac{6x}{y^2}$ B. $\displaystyle \frac{60xy}{10y^2}$ C. $\displaystyle \frac{6x^2}{y^2}$ D. $\displaystyle \frac{6x^2}{y}$ I did it like this: $\displaystyle \frac{4xy}{5}+\frac{15x}{2y^2}$ $\displaystyle =\frac{2y^2(4xy)}{10y^2}+\frac{5(15x)}{10y^2}$ $\displaystyle =\frac{8xy^3+75x}{10y^2}$ Got stuck there. Can it be simplified further?
Your answer looks good to me. (Well,, you can factor an x out in the numerator, but that's a little thing.) Pity that none of the given answers are correct. Are you certain there are no typos here?

-Dan

September 10th, 2014, 09:13 PM   #3
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 Originally Posted by Monox D. I-Fly 5. The simple form of $\displaystyle \frac{4xy}{5}+\frac{15x}{2y^2} is ....$
Evident that above should be: $\displaystyle \frac{4xy}{5}*\frac{15x}{2y^2}$
Then 6x^2 / 7 is solution.

September 11th, 2014, 03:53 AM   #4
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 Originally Posted by topsquark Your answer looks good to me. (Well,, you can factor an x out in the numerator, but that's a little thing.) Pity that none of the given answers are correct. Are you certain there are no typos here? -Dan
There are no typos in my side, however the post below yours makes sense:
Quote:
 Originally Posted by Denis Evident that above should be: $\displaystyle \frac{4xy}{5}*\frac{15x}{2y^2}$ Then 6x^2 / 7 is solution.
Maybe the one who wrote by handwriting intended to write $\displaystyle \times$ but the symbol was a bit slant so the typewriter mistook it as a +.

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