# How do you solve (s-3)/(s+4)=6/(s^2-16)?

Nov 8, 2016

$s - 3 = \frac{6 \left(s + 4\right)}{{s}^{2} - 16}$

Factor the denominator.

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

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

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

${s}^{2} - 3 s - 4 s + 12 = 6$

${s}^{2} - 7 s + 6 = 0$

$\left(s - 6\right) \left(s - 1\right) = 0$

$s = 6 \mathmr{and} 1$

Hopefully this helps!