# How do you solve 12/(a+3)+6/(a^2-9)=8/(a+3)?

Mar 27, 2018

$\textcolor{b l u e}{x = \frac{3}{2}}$

#### Explanation:

$\frac{12}{a + 3} + \frac{6}{{a}^{2} - 9} = \frac{8}{a + 3}$

factor $\left({a}^{2} - 9\right)$

This is the difference of two squares:

$\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

$\left({a}^{2} - {3}^{2}\right) = \left(a + 3\right) \left(a - 3\right)$

$\frac{12}{a + 3} + \frac{6}{\left(a + 3\right) \left(a - 3\right)} = \frac{8}{a + 3}$

Multiply through by $\left(x + 3\right)$

$\left(a + 3\right) \frac{12}{a + 3} + \left(a + 3\right) \frac{6}{\left(a + 3\right) \left(a - 3\right)} = \left(a + 3\right) \frac{8}{a + 3}$

Cancel:

$\cancel{\left(a + 3\right)} \frac{12}{\cancel{\left(a + 3\right)}} + \cancel{\left(a + 3\right)} \frac{6}{\cancel{\left(a + 3\right)} \left(a - 3\right)} = \cancel{\left(a + 3\right)} \frac{8}{\cancel{\left(a + 3\right)}}$

$12 + \frac{6}{x - 3} = 8$

Subtract $12$

$\frac{6}{x - 3} = - 4$

Multiply by $\left(x - 3\right)$

$\left(x - 3\right) \frac{6}{x - 3} = - 4 \left(x - 3\right)$

Cancel:

$\cancel{\left(x - 3\right)} \frac{6}{\cancel{x - 3}} = - 4 \left(x - 3\right)$

$6 = - 4 \left(x - 3\right)$

Divide by $- 4$

$x - 3 = \frac{6}{-} 4$

$x = \frac{6}{-} 4 + 3 = \frac{3}{2}$

$\textcolor{b l u e}{x = \frac{3}{2}}$