# Solve: x/2x-6 - 3/x^2 - 6x + 9 = x - 2/3x - 9

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You need to solve the following equation, such that:

`x/(2x - 6) - 3/(x^2 - 6x + 9) = (x - 2)/(3x - 9)`

You need to convert each denominator into its factored form such that:

`x/(2(x - 3)) - 3/((x - 3)^2) = (x - 2)/(3(x - 3))`

You need to bring the fractions to a common denominator, such that:

`((3x(x - 3)) - 6*3)/(6(x - 3)^2) = (2(x - 3))/(6(x - 3)^2)`

You need to reduce duplicate factors, such that:

`3x(x - 3) - 18 = 2(x - 3)`

`3x^2 - 9x - 18 - 2x + 6 = 0 => 3x^2 - 11x - 12 = 0`

You need to use quadratic formula to evaluate the solutions to quadratic equation `ax^2 + bx + c = 0` , such that:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

Identifying `a = 3, b = -11, c = -12` , yields:

`x_(1,2) = (11+-sqrt(121 + 144))/6 => x_(1,2) = (11+-sqrt265)/6`

**Hence, evaluating the solutions to the given equation, yields **`x_1 = (11+sqrt265)/6, x_2 = (11-sqrt265)/6.`