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

$\dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}$

$\dfrac{1}{x - 1} - \dfrac{1}{x} = \dfrac{1}{x + 3} - \dfrac{1}{x + 4}\\[1em]$

Since L.C.M. of denominators x and (x - 1) = x(x - 1) and L.C.M. of (x + 3) and (x + 4) = (x + 3)(x + 4),

$⇒ \dfrac{1 \times x}{x(x - 1)} - \dfrac{1 \times (x - 1)}{x(x - 1)} = \dfrac{1 \times (x + 4)}{(x + 3) \times (x + 4)} - \dfrac{1 \times (x + 3)}{(x + 4) \times (x + 3)}\\[1em] ⇒ \dfrac{x}{x^2 - x} - \dfrac{x - 1}{x^2 - x} = \dfrac{x + 4}{x^2 + 3x + 4x + 12} - \dfrac{x + 3}{x^2 + 3x + 4x + 12}\\[1em] ⇒ \dfrac{x - (x - 1)}{x^2 - x} = \dfrac{x + 4}{x^2 + 7x + 12} - \dfrac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{x - x + 1}{x^2 - x} = \dfrac{x + 4}{x^2 + 7x + 12} - \dfrac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{x - x + 1}{x^2 - x} = \dfrac{(x + 4) - (x + 3)}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{1}{x^2 - x} = \dfrac{x + 4 - x - 3}{x^2 + 7x + 12}\\[1em] ⇒ \dfrac{1}{x^2 - x} = \dfrac{1}{x^2 + 7x + 12}\\[1em]$

By cross multiplying, we get :

$⇒ x^2 + 7x + 12 = x^2 - x\\[1em] ⇒ x^2 - x^2 + 7x + x = - 12\\[1em] ⇒ 8x = - 12\\[1em] ⇒ x = - \dfrac{12}{8}\\[1em] ⇒ x = - \dfrac{3}{2}\\[1em] ⇒ x = - 1\dfrac{1}{2}$

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