# Thread: Addition and subtraction with fractions Factorization

1. ## Addition and subtraction with fractions Factorization

I used Google Translate for the title because my English isn't that good...

Anyway,
What am I doing wrong?
Here's my attempt to solve this exercise:

(http://img52.imageshack.us/img52/1108/question.png)

And here's the solution:

(http://img21.imageshack.us/img21/6429/answerz.png)

Can anyone explain me how to solve that right?
Thanks!

2. your mistake is in the numerator of the 3rd expression ...

$\displaystyle \frac{(a-b)(a+b) + 3ab + b^2 - 4ab}{3b(a-b)}$

$\displaystyle \frac{a^2 - b^{\textcolor{red}{2}} + 3ab + b^2 - 4ab}{3b(a-b)}$

$\displaystyle \frac{a^2 - ab}{3b(a-b)}$

$\displaystyle \frac{a(a - b)}{3b(a-b)}$

$\displaystyle \frac{a}{3b}$

3. You did a mistake: $\displaystyle (a-b)(a+b)=a^2-b^2$ not $\displaystyle a^2-b$.

4. Thanks!!!

5. The first couple of steps are correct. You have
$\displaystyle \frac{a+b}{3b}+ \frac{a}{a-b}+ \frac{b-4a}{3a-3b}$ and correctly recognize that the "3" in that last denominator can be factored out:
$\displaystyle \frac{a+b}{3b}+ \frac{a}{a-b}+ \frac{b-4a}{3(a-b)}$
so the common denominator is 3b(a-b).

Multiplying both numerator and denominator of: the first fraction by (a-b); the second fraction by 3b; the third fraction by b gives

$\displaystyle \frac{(a+b)(a-b)+ a(3b)+ (b- 4a)(b)}{3b(a-b)}$

But now you have $\displaystyle a^2- b$ as the first two terms in the next fraction and (a+b)(a-b)= $\displaystyle a^2- b^2$. That was probably a typo but you continued it through your answer. You should have
$\displaystyle \frac{a^2- b^2+ 3ab+ b^2- 4ab}{3b(a-b)}$.

Blast! Skeeter beat me to it again!

Now the "$\displaystyle b^2$" terms cancel and the fraction simplifies.

### Addition and subtraction if fractions in factorisation

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