Algebraic addition/simplification prob.

**jharrisbr8** · May 13th 2011, 05:05 AM

Simplify,

$\frac{a}{ab+{b}^{2 }}+\frac{b}{{a}^{2}+ab}$

I get the principle i.e. multiply denominators to find common denominator, add, factor, simplify. I've done loads of examples but for some reason I just can't get this one. I know it's relatively straight forward, I think my problems somewhere in the simplification/indices area.

I get as far as:-

$=\frac{(a)({a}^{2}+ab) + (b)(ab+{b}^{2})}{(ab+{b}^{2})({a}^{2}+ab)}$

$=\frac{({a}^{3}+{a}^{2}b)+(a{b}^{2}+{b}^{3})}{(ab+ {b}^{2})({a}^{2}+ab)}$

At this point it all seems to get overly complicated.
Anybody break it down a bit for me?

**Ackbeet** · May 13th 2011, 05:09 AM

Always look to see if you can simplify before plowing ahead with fractions. You didn't do anything wrong, but you're making life a bit harder for yourself than you need to. Here's what I would do:

$\frac{a}{ab+b^{2}}+\frac{b}{a^{2}+ab}=\frac{a}{b(a +b)}+\frac{b}{a(a+b)}.$

Does that make things a bit easier?

**jharrisbr8** · May 13th 2011, 05:56 AM

Thanks, this was one of the many roads I started down, but never quite got to the end of. I can at least, get to the answer now,

$\frac{{a}^{2}+{b}^{2}}{ab(a+b)}$

but remain a bit unconvinced by my methodology.

Am I right in thinking

$b(a+b) + a(a+b)$ = A common denominator $ab(a+b)$

In which case I only need to muliply each numerator by the 'missing' bit of the denominator, i.e.

$\frac{a}{b(a+b)}$ x $\frac{a}{a} = \frac{{a}^{2} }{ab(a+b)}$

and the same again for the other half?

**Ackbeet** · May 13th 2011, 06:01 AM

Exactly right. Be convinced!

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