Question #e806e

Aug 19, 2015

$\frac{2 x - 3}{2 x - 5}$

Explanation:

Your starting expression looks like this

$\frac{2 x + \frac{x}{x - 2}}{2 x - \frac{x}{x - 2}}$

This can be rewritten as

$\left(2 x + \frac{x}{x - 2}\right) \cdot \frac{1}{\left(2 x - \frac{x}{x - 2}\right)}$

Find the common denominator for the two terms that are written in parantheses

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

and

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

$\frac{2 x \left(x - 2\right) + x}{\left(x - 2\right)} \cdot \frac{1}{\frac{2 x \left(x - 2\right) - x}{x - 2}}$

which is equivalent to

$\frac{2 x \left(x - 2\right) + x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}{2 x \left(x - 2\right) - x}$

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

Expand the parantheses to g et

$\frac{2 x \cdot x + 2 x \cdot \left(- 2\right) + x}{2 x \cdot x + 2 x \cdot \left(- 2\right) - x}$

$\frac{2 {x}^{2} - 4 x + x}{2 {x}^{2} - 4 x - x} = \frac{2 {x}^{2} - 3 x}{2 {x}^{2} - 5 x}$

You can simplify this further by dividing the numerator and denominator by $x$ to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \left(2 x - 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \left(2 x - 5\right)} = \textcolor{g r e e n}{\frac{2 x - 3}{2 x - 5}}$