Algebra 2/Solve for x.?

x / (3x + 6) + 2 / (3x - 6) = 1 / (x^2 - 4). Factor each denominator.

x / 3(x + 2) + 2 / 3(x - 2) = 1 / (x + 2)(x - 2). Your common denominator is 3(x + 2)(x - 2). By multiplying all terms by the common denominator the common denominators will cancel out and you will be left with:

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

x^2 - 2x + 2x + 4 = 3. The - 2x and + 2x cancel each other out.

x^2 + 4 = 3. Subtract 4 from both sides.

x^2 = - 1. Take the square root of each side.

x = + or - (sq rt - 1)

x / 3x+6 + 2 / 3x-6 = 1 / X^2 - 4

[x(3x - 6) + 2 (3x + 6)] over 9x^2 - 36 = 1/ x^2 -4

[3x^2 - 6x + 6x + 12 ] over 9 (x^2 - 4) = 1/ x^2 - 4

[3x^2 + 12] over 9(x^2 - 4) = 1/ x^2 - 4 (multiply 9(x^2 - 4) on both sides)

3 (x^2 + 4) = 9

x^2 + 4 = 3

x^2 = -1

x = √-1

x = i

The gcf is 3(x+2)(x-2)

Multiply everything by that and get

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

x^2 - 2x + 2x + 4 = 3

x^2 = -1

x = √-1

x = + or - i