# Solve `3/(x^2-2x-8) + 1/(x+2) = 4/(x^2-16)`

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### 1 Answer

The equation `3/(x^2-2x-8) + 1/(x+2) = 4/(x^2-16)` has to be solved.

`3/(x^2-2x-8) + 1/(x+2) = 4/(x^2-16)`

=> `3/(x^2-4x + 2x-8) + 1/(x+2) = 4/(x^2-16)`

=> `3/(x(x - 4) + 2(x - 4)) + 1/(x+2) = 4/(x^2-16)`

=> `3/((x + 2)(x - 4)) + 1/(x+2) = 4/((x - 4)(x + 4))`

Let x + 2 = y

=> `3/(y*(y - 6)) + 1/y = 4/((y - 6)(y + 2))`

=> `(3*(y+2))/(y*(y - 6)(y+2)) + ((y - 6)(y+2))/(y(y - 6)(y+2)) = (4y)/(y(y - 6)(y + 2))`

=> `(3*(y+2)) + ((y - 6)(y+2)) = 4y`

=> 3y + 6 + y^2 - 4y - 12 = 4y

=> y^2 -5y - 6 = 0

=> y^2 - 6y + y - 6 = 0

=> y(y - 6) + 1(y - 6) = 0

=> (y + 1)(y - 6) = 0

substituting y = x + 2

=> (x + 3)(x - 4) = 0

=> x = -3 and x = 4

x = 4 is not a solution of the original equation as (x - 4) is a term in the denominator

**The solution of the equation `3/(x^2-2x-8) + 1/(x+2) = 4/(x^2-16)` is x = -3 **