Solve the following equation by factorization:

$\dfrac{3x + 1}{7x + 1} = \dfrac{5x + 1}{7x + 5}$

Given,

$\Rightarrow \dfrac{3x + 1}{7x + 1} = \dfrac{5x + 1}{7x + 5}$

⇒ (3x + 1)(7x + 5) = (5x + 1)(7x + 1)

⇒ (21x² + 15x + 7x + 5) = (35x² + 5x + 7x + 1)

⇒ (21x² + 22x + 5) = (35x² + 12x + 1)

⇒ (35x² + 12x + 1) - (21x² + 22x + 5) = 0

⇒ 35x² + 12x + 1 - 21x² - 22x - 5 = 0

⇒ 14x² - 10x - 4 = 0

⇒ 2(7x² - 5x - 2) = 0

⇒ 7x² - 5x - 2 = 0

⇒ 7x² - 7x + 2x - 2 = 0

⇒ 7x(x - 1) + 2(x - 1) = 0

⇒ (7x + 2)(x - 1) = 0

⇒ (7x + 2) = 0 or (x - 1) = 0     [Using Zero-product rule]

⇒ 7x = -2 or x = 1

⇒ x = $\dfrac{-2}{7}$ or x = 1.

Hence, $x = \Big{1, \dfrac{-2}{7}\Big}$.