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 October 29th, 2009, 04:56 AM #1 Newbie   Joined: Sep 2009 Posts: 8 Thanks: 0 Simplifying Complex Fractions How do you simplify this complex fraction? 1/a + 1/b over ab? The answer given is b + a over a^2 times b^2 but I don't know how this was derived and I would like to see a step by step explanation if anyone could be so kind. Thanks and cheers.
 October 29th, 2009, 05:25 AM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Simplifying Complex Fractions If you are wanting to simplify this fraction: $\frac{\frac{1}{a} + \frac{1}{b}}{ab}$, here's how: These two are the same: $\frac{\frac{1}{a} + \frac{1}{b}}{ab} \mbox{ and } \frac{\frac{1}{a} + \frac{1}{b}}{\frac{ab}{1}}$ Combine the numerator into one fraction, using the least common multiple: $\frac{1}{a} + \frac{1}{b}= \frac{1}{a}\cdot\frac{b}{b} + \frac{1}{b}\cdot\frac{a}{a} = \frac{b}{ab} + \frac{a}{ab} = \frac{a + b}{ab}$ Now substitute this in for the numerator: $\frac{\frac{1}{a} + \frac{1}{b}}{\frac{ab}{1}}= \frac{\frac{a + b}{ab}}{\frac{ab}{1}}$ To divide by a fraction, you invert and multiply, so $\frac{\frac{a + b}{ab}}{\frac{ab}{1}}= \frac{a + b}{ab}\cdot\frac{1}{ab} = \frac{a + b}{a^2b^2}$
 October 29th, 2009, 05:41 AM #3 Newbie   Joined: Sep 2009 Posts: 8 Thanks: 0 Re: Simplifying Complex Fractions Wow, thanks. Perfectly clear and fast on the reply to boot. Really appreciate it.
 October 29th, 2009, 04:06 PM #4 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: Simplifying Complex Fractions Hello, Tyster! A complex fraction is one which has more than two "levels". To simplify one, multiply numerator and denominator by the LCD of the denominators. $\text{Example: }\;\frac{\frac{1}{2b} \,-\, \frac{1}{3a}} {\frac{1}{4b^2} \,-\, \frac{1}{9a^2}}$ $\text{The denominators are: }\:2b,\:3a,\:4b^2,\:9a^2$ [color=beige]. . [/color]$\text{The LCD is: }\:36a^2b^2$ $\text{Multiply top and bottom by }36a^2b^2:\;\;\; \frac{36a^2b^2\left(\frac{1}{2b} \,-\, \frac{1}{3x}\right)} {36a^2b^2\left(\frac{1}{4b^2} \,-\, \frac{1}{9a^2}\right)} \;=\;\frac{18a^2b \,-\,12ab^2}{9a^2\,-\,4b^2}$ [color=beige]. . [/color]and we've eliminated the messy part! The rest is just "bookkeeping" . . . factor and reduce, if possible. $\text{Factor: }\;\frac{6ab(3a\,-\,2b)}{(3a\,-\,2b)(3a\,+\,2b)}$ $\text{Reduce: }\;\frac{6ab}{3a\,+\,2b}$
 November 3rd, 2009, 10:23 AM #5 Member   Joined: Oct 2009 Posts: 32 Thanks: 0 Re: Simplifying Complex Fractions I would multiply top and bottom of the complex fraction by ab yielding (a + b)/(a)^2(b)^2 which simplifies to 1/a(b)^2 + 1/b(a)^2.
 November 3rd, 2009, 01:47 PM #6 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 Re: Simplifying Complex Fractions Since we are STILL doing homework here, I'd factor the bottom and simplify first. It's easier than first finding a LCD.

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