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

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Answer 1 · 2016-11-08T00:52:29

$s - 3= (6(s + 4))/(s^2 - 16)$

Factor the denominator.

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

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

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

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

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

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

$s = 6 and 1$

Hopefully this helps!

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