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
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