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

$\dfrac{3x}{x + 6} - \dfrac{x}{x + 5} = 2$

$\dfrac{3x}{x + 6} - \dfrac{x}{x + 5} = 2\\[1em]$

Since L.C.M. of denominators (x + 6) and (x + 5) = (x + 6)(x + 5), multiply each term with (x + 6)(x + 5) to get:

$⇒ \dfrac{3x(x + 5)(x + 6)}{(x + 6)} - \dfrac{x(x + 5)(x + 6)}{(x + 5)} = 2(x + 5)(x + 6)$

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

⇒ (3x² + 15x) - (x² + 6x) = 2(x² + 6x + 5x + 30)

⇒ 3x² + 15x - x² - 6x = 2(x² + 11x + 30)

⇒ 2x² + 9x = 2x² + 22x + 60

⇒ 2x² - 2x² + 9x - 22x = 60

⇒ -13x = 60

⇒ x = - $\dfrac{60}{13}$

⇒ x = - $4\dfrac{8}{13}$

Hence, the value of x is $- 4\dfrac{8}{13}$.