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Solve the following equation for $x$ : $\frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x}, x \neq 0, 1, 2$
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$\frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \implies x[x - 1 + 2(x - 2)] = 6(x - 1)(x - 2) \implies 3x^2 - 13x + 12 = 0 \implies (3x - 4)(x - 3) = 0 \implies x = 3, x = \frac{4}{3}$