# How do you solve (9/x)+9/(x-2) = 12?

Dec 5, 2016

$x = \frac{1}{2}$ and $x = 3$

#### Explanation:

First, you can get each fraction over a common denominator, in this case $x \left(x - 2\right)$, so the fractions can be added:

$\left(\frac{x - 2}{x - 2}\right) \left(\frac{9}{x}\right) + \left(\frac{x}{x}\right) \left(\frac{9}{x - 2}\right) = 12$

$\frac{9 \left(x - 2\right) + 9 x}{x \left(x - 2\right)} = 12$

We can now expand and add the numerator:

$\frac{9 x - 18 + 9 x}{x \left(x - 2\right)} = 12$

$\frac{18 x - 18}{x \left(x - 2\right)} = 12$

We can now eliminate the fraction through multiplication on each side of the equation to keep the equation balanced:

$\frac{18 x - 18}{{x}^{2} - 2 x} = 12$

$\left({x}^{2} - 2 x\right) \frac{18 x - 18}{{x}^{2} - 2 x} = 12 \left({x}^{2} - 2 x\right)$

$\cancel{\left({x}^{2} - 2 x\right)} \frac{18 x - 18}{\cancel{\left({x}^{2} - 2 x\right)}} = 12 {x}^{2} - 24 x$

$18 x - 18 = 12 {x}^{2} - 24 x$

We can now make a single quadratic and factor:

$18 x - 18 - 18 x + 18 = 12 {x}^{2} - 24 x - 18 x + 18$

$12 {x}^{2} - 42 x + 18 = 0$

$\left(6 x - 3\right) \left(2 x - 6\right) = 0$

We can now solve each factor for $0$:

$6 x - 3 = 0$

$6 x - 3 + 3 = 0 + 3$

$6 x = 3$

$\frac{6 x}{6} = \frac{3}{6}$

$x = \frac{1}{2}$

and

$2 x - 6 = 0$

$2 x - 6 + 6 = 0 + 6$

$2 x = 6$

$\frac{2 x}{2} = \frac{6}{2}$

$x = 3$