# [SOLVED] Simplification of Algeraic Fractions

• May 4th 2009, 04:23 AM
waven
[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
• May 4th 2009, 04:42 AM
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:

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

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

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

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

(Wink)
• May 4th 2009, 05:00 AM
waven
Quote:

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:

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

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

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

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

(Wink)

(Surprised)... but is it still possible to simplify $\displaystyle \frac {3xy(x - 2)}{2xy(x - 2y)}$ further? like maybe $\displaystyle \frac {3(x - 2)}{2(x - 2y)}$
• May 4th 2009, 05:28 AM
e^(i*pi)
Quote:

Originally Posted by waven
(Surprised)... 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