Solve the following system of simultaneous linear equations by cross-multiplication method :

x + 3y + 4 = 0

3x - y = -2

Given equations can be written as,

x + 3y + 4 = 0

3x - y + 2 = 0

By cross multiplication method,

$\therefore \dfrac{x}{3 \times 2 - (-1) \times 4} = \dfrac{y}{4 \times 3 - 2 \times 1} = \dfrac{1}{1 \times (-1) - 3 \times 3}\\[1em] \Rightarrow \dfrac{x}{6 + 4} = \dfrac{y}{12 - 2} = \dfrac{1}{-1 - 9}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow \dfrac{x}{10} = \dfrac{1}{-10} \text{ and }\dfrac{y}{10} = \dfrac{1}{-10}\\[1em] \Rightarrow x = -\dfrac{10}{10} \text{ and }y = -\dfrac{10}{10}\\[1em] \Rightarrow x = -1 \text{ and }y = -1$

Hence, x = -1 and y = -1.