Aganippe76

2022-07-02

Simplify $\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}$
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 $\left(x-6\right)$ but I am getting ${x}^{2}-3x-18$? Any ideas?

trantegisis

$\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}=\frac{x+3}{x-6-2}\cdot \frac{1}{\frac{x+3}{x-6+8}}=\frac{x+3}{x-8}\cdot \frac{1}{\frac{x+3}{x+2}}=\frac{x+3}{x-8}\cdot \frac{x+2}{x+3}=\frac{x+2}{x-8}.$

Janet Forbes

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:
$\frac{\frac{x+3}{x-6-2}}{\frac{x+3}{x-6+8}}=\frac{x+3}{x-6-2}\cdot {\left(\frac{x+3}{x-6+8}\right)}^{-1}=\frac{x+3}{x-6-2}\cdot \frac{x-6+8}{x+3}=\frac{\left(x+3\right)\left(x-6+8\right)}{\left(x-6-2\right)\left(x+3\right)}$
Written like this, it is easier to see that the $x+3$ terms cancel:
$=\frac{x-6+8}{x-6-2}=\frac{x+2}{x-8}$

Do you have a similar question?