Solve the equation and find all solutions.?

Multiply all terms by the LCD: 2(x+1). Then solve.Combine like terms and factorise : x^2+x-12= (x+4)(x-3) x= -4 , 3.

(3/[x + 1]) - ([x - 2]/2) = (x - 2)/(x + 1)
3 - ([{x - 2}{x + 1}]/2) = x - 2
(3/[x - 2]) - ([x + 1]/2) = 1
3/(x - 2) = 1 + ([x + 1]/2)
3/(x - 2) = (2 + x + 1)/2
6 = (x - 2)(x + 3)
x² + 3x - 2x - 6 = 6
x² + x - 12 = 0
(x + 4)(x - 3) = 0

Values of x:
x = - 4, x = 3

Answer: x = - 4, 3

Proof (x = - 4):
(3/[- 4 + 1]) - ([- 4 - 2]/2) = (- 4 - 2)/(- 4 + 1)
(3/- 3) - (- 6/2) = - 6/- 3
- 1 + 3 = 2
2 = 2

Proof (x = 3):
(3/[ 3 + 1]) - ([3 - 2]/2) = (3 - 2)/(3 + 1)
3/4 - (1/2) = 1/4
3/4 - 2/4 = 1/4
1/4 = 1/4

3/(x + 1) - (x - 2)/2 = (x - 2)/(x + 1)
2[x + 1][3/(x + 1) - (x - 2)/2] = 2[x + 1][(x - 2)/(x + 1)]
6 - (x + 1)(x - 2) = 2(x - 2)
6 - x^2 - x + 2x + 2 = 2x - 4
x^2 + x - 2x + 2x - 6 - 2 - 4 = 0
x^2 + x - 12 = 0
x^2 + 4x - 3x - 12 = 0
(x^2 + 4x) - (3x + 12) = 0
x(x + 4) - 3(x + 4) = 0
(x + 4)(x - 3) = 0

x + 4 = 0
x = -4

x - 3 = 0
x = 3

∴ x = -4 , 3

6 - (x - 2)(x + 1) = 2 (x - 2)
6 - (x² - x - 2) = 2 x - 4
6 - x² + x + 2 = 2x - 4
8 - x² + x = 2x - 4
0 = x² + x - 12
(x + 4)(x - 3) = 0
x = - 4 , x = 3

you have to multiply all denominators out
so multiply both sides by (x+1)
then by 2

so you'll have 6 - (x+1)(x-2) = (x-2)(x+1)
or 6 - (x+1)(x-2) = (x+1)(x-2)
so add (x+1)(x-2) to both sides
6 = 2(x+1)(x-2)
6 = 2(x^2 -2x +x -2) by multiplying the brackets together
6 = 2x^2 -4x +2x -4 multiplying by the 2 out the front
6 = 2x^2 -2x -4 simplifying
0 = 2x^2 -2x -10 make it equal zero
0 = 2(x^2 -x -5) common factor of 2

and so x^2 -x -5 = 0, find the values of x that make it equal zero

or x^2 -x = 5

i cant be bothered with that part but there ya go