$2 a x + 3 b y = (a + 2 b), 3 a x + 2 b y = (2 a + b) .$

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Solution

The given equations may be written as:
2ax + 3by − (a + 2b) = 0 ...(i)
3ax + 2by − (2a + b) = 0 ...(ii)
Here, a₁= 2a, b₁ = 3b, c₁ = −(a + 2b), a₂ = 3a, b₂ = 2b and c₂ = −(2a + b)
By cross multiplication, we have:

∴ $\frac{x}{[3 b \times (- (2 a + b)) - 2 b \times (- (a + 2 b))]} = \frac{y}{[- (a + 2 b) \times 3 a - 2 a \times (- (2 a + b))]} = \frac{1}{[2 a \times 2 b - 3 a \times 3 b]}$
⇒ $\frac{x}{(- 6 a b - 3 b^{2} + 2 a b + 4 b^{2})} = \frac{y}{(- 3 a^{2} - 6 a b + 4 a^{2} + 2 a b)} = \frac{1}{4 a b - 9 a b}$
⇒ $\frac{x}{b^{2} - 4 a b} = \frac{y}{a^{2} - 4 a b} = \frac{1}{- 5 a b}$
⇒ $\frac{x}{- b (4 a - b)} = \frac{y}{- a (4 b - a)} = \frac{1}{- 5 a b}$
⇒ $x = \frac{- b (4 a - b)}{- 5 a b} = \frac{(4 a - b)}{5 a}, y = \frac{- a (4 b - a)}{- 5 a b} = \frac{(4 b - a)}{5 b}$
Hence, $x = \frac{(4 a - b)}{5 a}$ and $y = \frac{(4 b - a)}{5 b}$ is the required solution.

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