# How do you solve  7/(u - 3) - 42/(u^2 - 9) =1 and find any extraneous solutions?

Feb 3, 2018

Roots are $\textcolor{g r e e n}{u = 3 , 4}$

#### Explanation:

$\frac{7}{u - 3} - \frac{42}{{u}^{2} - 9} = 1$

Taking LCM fo LHS as $\left({u}^{2} - 9\right)$,

$\left(7 \left(u + 3\right) - 42\right) = 1 \cdot \left({u}^{2} - 9\right)$

$7 u + 21 - 42 = {u}^{2} - 9$

${u}^{2} - 7 u - 21 + 42 - 9 = 0$

${u}^{2} - 7 u + 12 = 0$

${u}^{2} - 3 u - 4 u + 12 = 0$

$u \left(u - 3\right) - 4 \left(u - 3\right) = 0$

$\left(u - 3\right) \cdot \left(u - 4\right) = 0$

Roots are $\textcolor{g r e e n}{u = 3 , 4}$