maths - algebra fractions?

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

multiply eqn by (x+3)(1-3x)(x-2) to remove all denominators

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

remove denominator, by canceling out similar numerator

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

expand:

3[x - 3x^2 + 6x - 2] - 2 [x^2 + 3x - 2x - 6] = 2[x - 3x^2 + 3 - 9x]

simplify

3 [-3x^2 + 7x -2] - 2 [x^2 +x - 6] = 2 [-3x^2 - 8x + 3]

-9x^2 + 21x - 6 - 2x^2 - 2x + 12 = -6x^2 - 16x + 6

move everything to the left

-9x^2 - 2x^2 + 6x^2 + 21x - 2x + 16x - 6 + 12 - 6 = 0

-5 x^2 + 35x = 0

divide by 5

-x^2 + 7x = 0

x (-x +7) = 0

x = 0 or

-x + 7 = 0 ==> x = 7

See my Workings I will try to make it clear as I can !!!

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

Or, {3(1 - 3x) - 2(x + 3)}/(x + 3)(1 - 3x) = 2/(x - 2)

Or, {3 - 9x - 2x -6}/(x + 3)(1 - 3x) = 2/(x - 2)

Or, (x - 2)(-11x -3) = 2(x + 3)(1 - 3x)

Or, {-11x^2 -3x + 22x + 6} = 2x - 6x^2 + 6 - 18x

Or, -11x^2 + 6x^2 + 19x - 2x + 18x = 0

Or, -5x^2 + 35x = 0

Or, 5x^2 - 35x = 0

Or, 5x (x -7) = 0

Or, 5x = 0 Or, x - 7 = 0

This gives x = 0 or 7

I hope you got my workings !!

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