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

ax + by = (a − b), bx − ay = (a + b)

Given,

Equations:

⇒ ax + by - (a − b) = 0

⇒ bx − ay - (a + b) = 0

By cross-multiplication method,

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

Hence, x = 1 and y = -1.