How do you solve $\frac{s - 3}{s + 4} = \frac{6}{s^{2} - 16}$ $(s-3)/(s+4)=6/(s^2-16)$ ?

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Answer 1 · 2016-11-08T00:52:29

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

Factor the denominator.

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

$s - 3 = \frac{6}{s - 4}$ $s - 3 = 6/(s - 4)$

$(s - 3) (s - 4) = 6$ $(s- 3)(s - 4) = 6$

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

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

$(s - 6) (s - 1) = 0$ $(s - 6)(s - 1) = 0$

$s = 6 and 1$ $s = 6 and 1$

Hopefully this helps!

How do you solve s−3s+4=6s2−16(s-3)/(s+4)=6/(s^2-16)?