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**Prove It** OK, for starters, don't jump to multiplying by the common denominator. I ALWAYS write the fractions with their common denominators first, it is clearer and reduces mistakes.

So $\displaystyle \displaystyle \frac{6\cdot 3}{6(x - 3)^2} + \frac{2(x - 3)(x - 2)}{6(x - 3)^2} = \frac{3x(x - 3)}{6(x - 3)^2}$

$\displaystyle \displaystyle \frac{18}{6(x - 3)^2} + \frac{2x^2 - 10x + 12}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x- 3)^2}$

$\displaystyle \displaystyle \frac{2x^2 - 10x + 30}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x - 3)^2}$

$\displaystyle \displaystyle 2x^2 - 10x + 30 = 3x^2 - 9x$.

Now solve the resulting quadratic.