# Simplify $\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}$

$$\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}$$

How would I simplify this complex fraction? I know what the answer is, but I am just not sure how they got there.

I have tried multiplying both sides by $(x-6)$ but I am getting $x^2-3x-18$? Any ideas?

• Without an equals sign, there's nothing to solve. – Unit Jul 29 '16 at 14:35
• Without a correct formatting, or at least parentheses, it's impossible to understand what's exactly in this expression. – Bernard Jul 29 '16 at 14:39
• @Unit That comes off as very snide. How about making a constructive criticism and telling him the right word to use is "simplify". – Gregory Grant Jul 29 '16 at 14:39
• @GregoryGrant: I have tried reading both sides of the question but couldn't figure out what the problem to be solved was. Any ideas? – user21820 Jul 29 '16 at 15:01
• @GregoryGrant: I think you didn't get my point. The asker mentioned "sides" but I don't see any sides! – user21820 Jul 29 '16 at 15:36

$$\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}=\frac{x+3}{x-6-2}\cdot\frac{1}{\frac{x+3}{x-6+8}}=\frac{x+3}{x-8}\cdot\frac{1}{\frac{x+3}{x+2}}=\frac{x+3}{x-8}\cdot\frac{x+2}{x+3}=\frac{x+2}{x-8}.$$
Note that dividing by a fraction is equivalent to multiplying by its reciprocal -- hence this might be the first thing you want to do, so that you only have one numerator and denominator: $$\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}} = \frac{x+3}{x-6-2}\cdot\left(\frac{x+3}{x-6+8}\right)^{-1} = \frac{x+3}{x-6-2} \cdot \frac{x-6+8}{x+3} = \frac{(x+3)(x-6+8)}{(x-6-2)(x+3)}$$
Written like this, it is easier to see that the $x+3$ terms cancel:
$$=\frac{x-6+8}{x-6-2}=\frac{x+2}{x-8}$$