# How do you subtract 6/(x+2)-(2x+3)/x?

Jan 14, 2017

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

#### Explanation:

Before subtracting we require the fractions to have a $\textcolor{b l u e}{\text{common denominator}}$

This is achieved as follows.

$\left(\frac{6}{x + 2} \times \frac{x}{x}\right) - \left(\frac{2 x + 3}{x} \times \frac{\left(x + 2\right)}{\left(x + 2\right)}\right)$

That is, multiply the numerator/denominator of the first fraction by the denominator of the second fraction and multiply the numerator/denominator of the second fraction by the denominator of the first fraction.

$= \frac{6 x}{x \left(x + 2\right)} - \frac{\left(2 x + 3\right) \left(x + 2\right)}{x \left(x + 2\right)}$

We now have a common denominator and can subtract the terms on the numerator.

$= \frac{6 x - \left(2 {x}^{2} + 7 x + 6\right)}{x \left(x + 2\right)}$

$= \frac{6 x - 2 {x}^{2} - 7 x - 6}{x \left(x + 2\right)} = \frac{- 2 {x}^{2} - x - 6}{x \left(x + 2\right)}$

Take out a common factor of - 1 on the numerator.

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