Simplifying Expression 3 Success

**Sooda** · July 9th 2009, 12:19 PM

I solved this by my own, using things learned from my previous two threads. I am so proud. I post it here so maybe others can learn from it if they decide to.

Expression to simplify:
$(\frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9})\div\frac{8}{4a^2+12a+9}$

Answer:
$\frac{2a+3}{2}$

Step by step:
$\frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9}=\frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{4a^2-9}$

${4a^2-9}$ is formula: $a^2-b^2=(a+b)(a-b)$
${4a^2-9}=(2a+3)(2a-3)$

$\frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{(2a+3)(2a-3)}$

$\frac{5(2a-3)-2(2a+3)+(2a+9)}{(2a+3)(2a-3)}$

$\frac{10a-15-4a-6+2a+9}{(2a+3)(2a-3)}=\frac{8a-12}{(2a+3)(2a-3)}$

Time to divide,
$4a^2+12a+9$ is formula: $a^2+2ab+b^2=(a+b)^2$
$4a^2+12a+9=(2a+3)^2$

$\frac{8a-12}{(2a+3)(2a-3)}\div\frac{8}{(2a+3)^2}$

$\frac{8a-12}{(2a+3)(2a-3)}*\frac{(2a+3)^2}{8}$

$\frac{8a-12}{(2a-3)}*\frac{(2a+3)}{8}$

$\frac{4(2a-3)}{(2a-3)}*\frac{(2a+3)}{8}$

$\frac{4}{1}*\frac{(2a+3)}{8}$

$\frac{2a+3}{2}$

So good to give something back.

**AlephZero** · July 9th 2009, 04:41 PM

Excellent job. Just one tip:

noticing that

$\frac{8a-12}{(2a+3)(2a-3)}=\frac{4(2a-3)}{(2a+3)(2a-3)}=\frac{4}{2a+3}$

would have saved you some trouble.

Cheers.

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