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!