# Thread: [SOLVED] Simplification of Algeraic Fractions

1. ## [SOLVED] Simplification of Algeraic Fractions

Q. Simplify the following:
$\displaystyle \frac {3x^2y - 6xy}{2x^2y - 4xy^2}$

i've got up to

$\displaystyle \frac {3xy(x - 2)}{2xy(x - 2y)}$

i tried simplifying it further but i just end up getting the wrong answer, which is $\displaystyle \frac{3}{2y}$
thanks

2. The answer you've been given is wrong. The only way for the expression to simplify that way would be for the original expression to change the power on the first of the "y" terms in the denominator:

. . . . .$\displaystyle \frac{3x^2y\, -\, 6xy}{2x^2y^2\, -\, 4xy^2}$

. . . . .$\displaystyle \frac{3xy(x\, -\, 2)}{2xy^2(x\, -\, 2)}$

. . . . .$\displaystyle \frac{3xy}{2xy^2}$

. . . . .$\displaystyle \frac{3}{2y}$

3. Originally Posted by stapel
The answer you've been given is wrong. The only way for the expression to simplify that way would be for the original expression to change the power on the first of the "y" terms in the denominator:

. . . . .$\displaystyle \frac{3x^2y\, -\, 6xy}{2x^2y^2\, -\, 4xy^2}$

. . . . .$\displaystyle \frac{3xy(x\, -\, 2)}{2xy^2(x\, -\, 2)}$

. . . . .$\displaystyle \frac{3xy}{2xy^2}$

. . . . .$\displaystyle \frac{3}{2y}$

... but is it still possible to simplify $\displaystyle \frac {3xy(x - 2)}{2xy(x - 2y)}$ further? like maybe $\displaystyle \frac {3(x - 2)}{2(x - 2y)}$

4. Originally Posted by waven
... but is it still possible to simplify $\displaystyle \frac {3xy(x - 2)}{2xy(x - 2y)}$ further? like maybe $\displaystyle \frac {3(x - 2)}{2(x - 2y)}$
Yes, that's correct - xy does cancel