Solve (question no. 2-22) for x :

$\dfrac{1}{x - 1} + \dfrac{2}{x - 2} = \dfrac{3}{x - 3}$

$\dfrac{1}{x - 1} + \dfrac{2}{x - 2} = \dfrac{3}{x - 3}$

Since L.C.M. of denominators (x - 1), (x - 2) and (x - 3) = (x - 1)(x - 2)(x - 3), multiply each term with (x - 1)(x - 2)(x - 3) to get :

$⇒ \dfrac{1(x - 1)(x - 2)(x - 3)}{(x - 1)} + \dfrac{2(x - 1)(x - 2)(x - 3)}{(x - 2)} = \dfrac{3(x - 1)(x - 2)(x - 3)}{x - 3}$

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

⇒ (x² - 2x - 3x + 6) + 2(x² - 1x - 3x + 3) = 3(x² - 1x - 2x + 2)

⇒ (x² - 5x + 6) + 2(x² - 4x + 3) = 3(x² - 3x + 2)

⇒ x² - 5x + 6 + 2x² - 8x + 6 = 3x² - 9x + 6

⇒ 3x² - 13x + 12 = 3x² - 9x + 6

⇒ 3x² - 3x² - 13x + 9x = 6 - 12

⇒ - 4x = - 6

⇒ x = $\dfrac{6}{4}$

⇒ x = $\dfrac{3}{2}$

⇒ x = $1\dfrac{1}{2}$

Hence, the value of x is $1\dfrac{1}{2}$.