Solve the following system of equations by using the method of cross multiplication:

x + 2y + 1 = 0, 2x − 3y = 12

Given,

Equations:

⇒ x + 2y + 1 = 0

⇒ 2x − 3y - 12 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(2) \times (-12) - (-3) \times (1)} = \dfrac{y}{(1) \times (2) - (-12) \times (1)} = \dfrac{1}{(1) \times (-3) - (2) \times (2)} \\[1em] \Rightarrow \dfrac{x}{(-24) + 3} = \dfrac{y}{2 + 12} = \dfrac{1}{-3 - 4} \\[1em] \Rightarrow \dfrac{x}{-21} = \dfrac{y}{14} = \dfrac{1}{-7} \\[1em] \Rightarrow \dfrac{x}{-21} = \dfrac{1}{-7} \text{ and } \dfrac{y}{14} = \dfrac{1}{-7} \\[1em] \Rightarrow x = \dfrac{-21}{-7} \text{ and } y = \dfrac{14}{-7} \\[1em] \Rightarrow x = 3 \text{ and } y = -2.$

Hence, x = 3 and y = -2.