solve 2x-1/x-5 > x+1/x+5?

(2x-1)/x-5 > (x+1)/(x+5)
Multiply both sides by (x-5)(x+5).

(2x-1)(x+5)/(x-5)(x+5) > (x+1)(x-5)/(x-5)(x+5)
(2x-1)(x+5)> (x+1)(x-5)
2x^2 + 10x -x -5> x^2 -5x +x -5
2x^2 + 9x - 5> x^2 - 4x -5
x^2 + 13x > 0
x(x + 13) > 0
x=0 or x=-13

I will solve (2x-1)/(x-5) > (x+1)/(x+5)

First note that x=5 or x=-5 would make a denominator equal to zero, so these are critical points that cannot be in the solution.

Now solve (2x-1)/(x-5) = (x+1)/(x+5).
Multiply both sides by (x-5)(x+5)
(2x-1)(x+5) = (x+1)(x-5)
Multiply on each side
2x²+9x-5 = x²-4x-5
x²+13x = 0
x(x+13) = 0
Thus x=0 and x=-13 are two more critical points.
Consider the five regions:
1. x<-13
2. -13<x<-5
3. -5<x<0
4. 0<x<5
5. 5<x

Try a point in each region to see if the original equation is true. (I tried the points -20, -10, -1, 1, and 6).
I found that in regions 1, 3, and 5 the original equation is true and in regions 2 and 4 the equation is false.

Hence the solution is all x such that
x<-13 or
-5<x<0 or
5<x