# 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?

##### 1 Answer

Aug 12, 2016

#### Answer:

#### Explanation:

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

The value

That leaves one solution