# How do you solve (10)/(x^2-2x)=4/x=5/(x-2)?

May 12, 2017

First, we must change the denominators to become the same
( common denominator )

The first thing to do is to expand all the denominators, to fully see the components of the fractions.
$\frac{10}{x \left(x - 2\right)} = \frac{4}{x} = \frac{5}{x - 2}$

If we multiply $\frac{4}{x}$ by $\frac{x - 2}{x - 2}$, and $\frac{5}{x - 2}$ by $\frac{x}{x}$, then all three fractions will have the common denominator of $x \left(x - 2\right)$:

$\frac{10}{x \left(x - 2\right)} = \frac{4 \left(x - 2\right)}{x \left(x - 2\right)} = \frac{5 \left(x\right)}{\left(x - 2\right) \left(x\right)}$

$\frac{10}{{x}^{2} - 2 x} = \frac{4 x - 8}{{x}^{2} - 2 x} = \frac{5 x}{{x}^{2} - 2 x}$

multiply all fractions by $\left({x}^{2} - 2 x\right)$

$\cancel{\left({x}^{2} - 2 x\right)} \times \frac{10}{\cancel{\left({x}^{2} - 2 x\right)}} = \cancel{\left({x}^{2} - 2 x\right)} \times \frac{4 x - 8}{\cancel{\left({x}^{2} - 2 x\right)}} = \cancel{\left({x}^{2} - 2 x\right)} \times \frac{5 x}{\cancel{\left({x}^{2} - 2 x\right)}}$

We are left with

$10 = \left(4 x - 8\right) = \left(5 x\right)$