# How do you simplify (2)/(x) + (2)/(x-1) - (2)/(x-2)?

Jun 19, 2018

See a solution process below:

#### Explanation:

To add fractions the fractions must be over a common denominator. First, we need to multiply each fraction by the appropriate form of $1$ to put them over a common denominator of:

$x \left(x - 1\right) \left(x - 2\right)$

$\frac{2}{x} + \frac{2}{x - 1} - \frac{2}{x - 2} \implies$

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

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

Next, we can expand the numerators for each fraction:

$\frac{\left(2 x - 2\right) \left(x - 2\right)}{x \left(x - 1\right) \left(x - 2\right)} + \frac{2 {x}^{2} - 4 x}{x \left(x - 1\right) \left(x - 2\right)} - \frac{2 {x}^{2} - 2 x}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{2 {x}^{2} - 4 x - 2 x + 4}{x \left(x - 1\right) \left(x - 2\right)} + \frac{2 {x}^{2} - 4 x}{x \left(x - 1\right) \left(x - 2\right)} - \frac{2 {x}^{2} - 2 x}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{2 {x}^{2} - 6 x + 4}{x \left(x - 1\right) \left(x - 2\right)} + \frac{2 {x}^{2} - 4 x}{x \left(x - 1\right) \left(x - 2\right)} - \frac{2 {x}^{2} - 2 x}{x \left(x - 1\right) \left(x - 2\right)}$

Then, we can add the numerators over the common denominator:

$\frac{\left(2 {x}^{2} - 6 x + 4\right) + \left(2 {x}^{2} - 4 x\right) - \left(2 {x}^{2} - 2 x\right)}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{\left(2 {x}^{2} - 6 x + 4 + 2 {x}^{2} - 4 x - 2 {x}^{2} + 2 x\right)}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{\left(2 {x}^{2} + 2 {x}^{2} - 2 {x}^{2} - 6 x - 4 x + 2 x + 4\right)}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{\left(2 + 2 - 2\right) {x}^{2} + \left(- 6 - 4 + 2\right) x + 4}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{2 {x}^{2} + \left(- 8\right) x + 4}{x \left(x - 1\right) \left(x - 2\right)} \implies$

$\frac{2 {x}^{2} - 8 x + 4}{x \left(x - 1\right) \left(x - 2\right)}$

If you need to simplify the denominator it would be:

$\frac{2 {x}^{2} - 8 x + 4}{\left({x}^{2} - x\right) \left(x - 2\right)} \implies$

$\frac{2 {x}^{2} - 8 x + 4}{{x}^{3} - 2 {x}^{2} - {x}^{2} + 2 x} \implies$

$\frac{2 {x}^{2} - 8 x + 4}{{x}^{3} + \left(- 2 - 1\right) {x}^{2} + 2 x} \implies$

$\frac{2 {x}^{2} - 8 x + 4}{{x}^{3} + \left(- 3\right) {x}^{2} + 2 x} \implies$

$\frac{2 {x}^{2} - 8 x + 4}{{x}^{3} - 3 {x}^{2} + 2 x}$