# How do I simplify (2x)/(x-3)-(x)/(x+3)?

Aug 9, 2015

$\frac{x \left(x + 9\right)}{\left(x - 3\right) \left(x + 3\right)}$

#### Explanation:

Think of how you simplify fractions without algebraic terms. (e.g. $\frac{1}{3} + \frac{2}{5}$. Determine the lowest common multiple between $3$ and $5$, which is $15$.

For the first fraction multiply $5$ on both numerator and denominator to get the denominator to $15$ and for the second fraction multiply $3$ on both numerator and denominator to get the denominator to $15$.

Thus, $\frac{1}{3} + \frac{2}{5} = \frac{5}{15} + \frac{6}{15} = \frac{11}{15}$

The principle for fractions with algebraic terms is the same. The questions asks to simplify $\frac{2 x}{x - 3} - \frac{x}{x + 3}$. Determine the lowest common multiple between $x - 3$ and $x + 3$, which is $\left(x - 3\right) \left(x + 3\right)$.

For the first fraction multiply $x + 3$ on both numerator and denominator to get the denominator to $\left(x - 3\right) \left(x + 3\right)$ and for the second fraction multiply $x - 3$ on both numerator and denominator to get the denominator to $\left(x - 3\right) \left(x + 3\right)$.

Thus,
(2x)/(x-3) - x/(x+3) =(2x(x+3))/((x-3)(x+3))-(x(x-3))/((x-3)(x+3)) =(2x^2+6x)/((x-3)(x+3))-(x^2-3x)/((x-3)(x+3)) =(2x^2+6x-x^2+3x)/((x-3)(x+3)) =(x^2+9x)/((x-3)(x+3)) =(x(x+9))/((x-3)(x+3))