# Thread: Simplify the complex fraction

1. ## Simplify the complex fraction

I can barely simplify regular fractions! My book's explanation of how to do this has me so terribly confused...

$\displaystyle \frac{\frac{b}{b+9}+\frac{7}{5b}}{\frac{b}{3b+27}+ \frac{4}{b}}$

2. Note that $\displaystyle 3b+27=3(b+9)$.
So the LCD is $\displaystyle 15b(b+9)$.

3. Originally Posted by Plato
Note that $\displaystyle 3b+27=3(b+9)$.
So the LCD is $\displaystyle 15b(b+9)$.
So, I would simply each individual fraction, then multiply both numerators by the LCD right?

4. Originally Posted by Fails_at_Math
So, I would simply each individual fraction, then multiply both numerators by the LCD right?
Right.

5. Originally Posted by Fails_at_Math
$\displaystyle \frac{\frac{b}{b+9}+\frac{7}{5b}}{\frac{b}{3b+27}+ \frac{4}{b}}$
Expanding a bit on the previous (and spot-on!) reply, multiply the fraction, top and bottom, by the LCM of the four sub-fractions:

$\displaystyle \frac{\displaystyle\frac{15b(b\, +\, 9)}{1}\left(\frac{b}{b\, +\, 9}\right)\, +\, \frac{15b(b\, +\, 9)}{1}\left(\frac{7}{5b}\right)}{\displaystyle\fra c{15b(b\, +\, 9)}{1}\left(\frac{b}{3(b\, +\, 9)}\right)\, +\, \frac{15b(b\, +\, 9)}{1}\left(\frac{4}{b}\right)}$

Cancel stuff to get:

$\displaystyle \frac{15b(b)\, +\, 3(b\, +\, 9)(7)}{5b(b)\, +\, 15(b\, +\, 9)(4)}$

I'll bet you can see where to go from there!

6. Now that I've finished my lunch...

$\displaystyle \frac{3(5b^2\, +\, 7b\, +\, 63)}{5(b^2\, +\, 12b\, +\, 108)}$

Should I leave it like this? Based on the answers of similar problems in the book, I think I should distribute it out to...

$\displaystyle \frac{15b^2\, +\, 21b\, +\, 189)}{5b^2\, +\, 60b\, +\, 540)}$

How's this look?

Much appreciate the help guys.