# How do you add or subtract 1/(x²-x) - 1/x?

Mar 14, 2018

The given expression simplifies to $\frac{- x + 2}{{x}^{2} - x}$

#### Explanation:

I will let the given expression be represented by $P$. So that

$P = \frac{1}{{x}^{2} - x} - \frac{1}{x}$

You need to manipulate (following the rules of algebra) the terms
$\frac{1}{{x}^{2} - x} \mathmr{and} \frac{1}{x}$
so that they have the same denominator.

Multiply top and bottom of $\frac{1}{{x}^{2} - x}$ by x. That yields

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

Multiply top and bottom of $\frac{1}{x}$ by (x^2 - x). That yields

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

Putting them together,

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

A few more simplification steps to go

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

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

So $P = \frac{- x + 2}{{x}^{2} - x}$ seems to be as far as simplification will take it.

I hope this helps,
Steve