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 31, 2016

Answer:

#{2}#.

Explanation:

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

Put on a common denominator.

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

Cancel the denominators and solve the resulting quadratic.

#x^2 + 3x + 4x + 12 + x^2 - 14x + 2x - 28 = x^2 + x - 8x - 8#

#x^2 + 2x - 8 = 0#

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

#x = -4 and 2#

However, #x = -4# is extraneous since it is one of the restrictions on the original equation (it renders the denominator #0#). The solution set is therefore #{2}#.

Hopefully this helps!