Simplify the complex fraction

**Fails_at_Math** · August 7th 2009, 09:01 AM

I can barely simplify regular fractions!

My book's explanation of how to do this has me so terribly confused...

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

Thanks for any help you can offer.

**Plato** · August 7th 2009, 09:07 AM

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

**Fails_at_Math** · August 7th 2009, 09:16 AM

Originally Posted by Plato

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

So, I would simply each individual fraction, then multiply both numerators by the LCD right?

**Plato** · August 7th 2009, 09:18 AM

Originally Posted by Fails_at_Math

So, I would simply each individual fraction, then multiply both numerators by the LCD right?

Right.

**stapel** · August 7th 2009, 09:22 AM

Originally Posted by Fails_at_Math

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

Expanding a bit on the previous (and spot-on!) reply, multiply the fraction, top and bottom, by the LCM of the four sub-fractions:

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

Cancel stuff to get:

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

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

**Fails_at_Math** · August 7th 2009, 10:30 AM

Now that I've finished my lunch...

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

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

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

How's this look?

Much appreciate the help guys.

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