Question

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?

Answer 1

`{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!