How do i work out this maths question

http://studymaths.co.uk/revise/32.html
Thanks in advance.

(2x + 3) / (x-4) - (2x - 8) / (2x+1) = 1 ... lots of algebra. Multiply by (x-4) and (2x+1) to clear the denominators, which gives

(2x+3)(2x+1) - (2x-8)(x-4) = (2x+1)(x-4) ... now multiply the binomials

4x^2 + 8x + 3 - [2x^2 -16x + 32] = 2x^2 - 7x - 4 ... now combine left side

2x^2 + 24x - 29 = 2x^2 - 7x - 4 ... combine again

31x - 25 = 0 ... solve by adding 25 to both sides, then divide by 31

x = 25/31

(2x + 3)/(x - 4) - (2x - 3)/(2x + 1) = 1
create common denominators:
(2x + 3)(2x + 1)/(x - 4)(2x + 1) - (2x - 3)(x - 4)/(x - 4)(2x + 1) = (x - 4)(2x + 1)/(x - 4)(2x + 1)
multiply through by (x - 4)(2x + 1):
(2x + 3)((2x + 1) - (2x - 3)(x - 4) = (x - 4)(2x + 1)
4x^2 + 2x + 6x + 3 - (2x^2 - 8x - 3x + 12) = 2x^2 + x - 8x - 4
4x^2 + 8x + 3 - 2x^2 + 11x - 12 = 2x^2 - 7x - 4
2x^2 + 19x - 9 = 2x^2 - 7x - 4
26x = 5
x = 5/26

- .--