# Thread: Another Algebraic Fraction

1. ## Another Algebraic Fraction

Simplify:

$\displaystyle \frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}$

2. Originally Posted by stellina_91
Simplify:

$\displaystyle \frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}$
Note that $\displaystyle x^2 - y^2 = (x-y)(x+y)$ ....

3. Same with this one.

4. Originally Posted by stellina_91
Same with this one.
Ditto.

5. $\displaystyle \frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}$

= $\displaystyle \frac{1}{x - y} + \frac{2 x - y}{(x + y)(x - y)}$

= $\displaystyle \frac{x^2 - y^2}{x - y} + \frac{(2 x - y)(x - y)}{(x + y)(x - y)}$

= $\displaystyle \frac{x^2 - y^2 + 2 x^2 - 2 x y - x y - y^2}{(x - y)(x + y)(x - y)}$

= $\displaystyle \frac{x^2 - y^2 + 2 x^2 - 3 x y - y^2}{(x - y)(x + y)(x - y)}$

= $\displaystyle \frac{3 x^2 - 2 y^2 - 3 x y}{(x - y)(x + y)(x - y)}$

6. Originally Posted by stellina_91
$\displaystyle \frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}$

= $\displaystyle \frac{1}{x - y} + \frac{2 x - y}{(x + y)(x - y)}$

= $\displaystyle \frac{x^2 - y^2}{x - y} + \frac{(2 x - y)(x - y)}{(x + y)(x - y)}$ Mr F says: No. The common denominator is (x + y)(x - y). The next line is:

$\displaystyle {\color{red}\frac{x + y}{(x - y)(x+y)} + \frac{2x - y}{(x + y)(x-y)}}$

[snip]
..

7. Why didn't you multiply 2x + y by x - y??

Thanks.

8. Originally Posted by stellina_91
Why didn't you multiply 2x + y by x - y??

Thanks.
I'm not sure what you're asking.

You get each fraction over a common denominator. The common denominator is (x + y)(x - y). Only one fraction needs to be re-written so that it is has the common denominator. I did that. The other fraction already has the common denominator - so nothing gets done to it.

9. It's ok, I understand now, thank for your help