can you solve this algebraic equation with fractions??!!?

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

We simplify as much as we can to get:

(-2/x) + 1 = (1/4)

We isolate the 'x' on one side to get:

-2/x = -3/4

We multiply by 'x' to get out of the fraction:

-2 = (-3x/4)

Multiply by 4 to get rid of all the fractions completely:

-8 = -3x

Divide by -3 to solve

8/3 = x

We plug back in to check and it comes out correct.

1/(x - 1) - 3/(x + 2) = 1/4
[1(x + 2)]/[(x + 2)(x - 1)] - [3(x - 1)]/[(x + 2)(x - 1)] = 1/4
(x + 2)/[(x + 2)(x - 1)] - (3x - 3)/[(x + 2)(x - 1)] = 1/4
(x + 2 - 3x + 3)/[(x + 2)(x - 1)] = 1/4
(-2x + 5) = [(x + 2)(x - 1)]/4
4(-2x + 5) = x^2 + 2x - x - 2
-8x + 20 = x^2 + x - 2
x^2 + x + 8x - 2 - 20 = 0
x^2 + 9x - 22 = 0
x^2 + 11x - 2x - 22 = 0
(x^2 + 11x) - (2x + 22) = 0
x(x + 11) - 2(x + 11) = 0
(x + 11)(x - 2) = 0

x + 11 = 0
x = -11

x - 2 = 0
x = 2

∴ x = -11 , 2

common denominator (x-1)(x+2)

((x+2) - 3(x-1))/(x-1)(x+2) = 1/4
(5 - 2x)/(x-1)(x+2) = 1/4
4(5-2x) = (x-1)(x+2)
20 - 8x = x^2 + x - 2
x^2 + 9x - 22 = 0
(x+11)(x-2)=0
x = -11 or 2