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

2bx + ay = 2ab

bx - ay = 4ab.

Given equations can be written as,

2bx + ay - 2ab = 0

bx - ay - 4ab = 0

By cross multiplication method,

$\therefore \dfrac{x}{a \times -(4ab) - (-a) \times [-2ab]} = \dfrac{y}{[-(2ab) \times b] - [-(4ab) \times 2b] } = \dfrac{1}{2b \times (-a) - b \times a} \\[1em] \Rightarrow \dfrac{x}{-4a^2b - 2a^2b} = \dfrac{y}{-2ab^2 + 8ab^2} = \dfrac{1}{-2ab - ab} \\[1em] \Rightarrow -\dfrac{x}{6a^2b} = \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \therefore -\dfrac{x}{6a^2b} = -\dfrac{1}{3ab} \text{ and } \dfrac{y}{6ab^2} = -\dfrac{1}{3ab} \\[1em] \Rightarrow x = \dfrac{6a^2b}{3ab} \text{ and } y = -\dfrac{6ab^2}{3ab} \\[1em] \Rightarrow x = 2a \text{ and } y = -2b.$

Hence, x = 2a and y = -2b.