How do you solve #(x+3)/(x^2-6x-16)+(x-14)/(x^2-4x-32)=(x+1)/(x^2+6x+8)# and check for extraneous solutions?

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Answer 1 · 2016-08-12T23:14:59

#x=2# (extraneous solution #x=-4#)

First factor all of the denominators and multiply through by the least common multiple:

#x^2-6x-16 = (x+2)(x-8)#

#x^2-4x-32 = (x-8)(x+4)#

#x^2+6x+8 = (x+2)(x+4)#

So the least common multiple of the denominators is:

#(x+2)(x+4)(x-8)#

Multiplying the original equation by this we get:

#(x+3)(x+4)+(x-14)(x+2)=(x+1)(x-8)#

Which expands out to:

#x^2+7x+12+x^2-12x-28=x^2-7x-8#

Subtract the right hand side from the left and simplify to get:

#0 = x^2+2x-8 = (x+4)(x-2)#

Hence zeros #x=-4# and #x=2#

The value #x=-4# is extraneous, since it makes two of the denominators of the original equation zero and division by #0# is undefined.

That leaves one solution #x=2#