(√3 + √2)x - (√2 - √5)y = 0 -----(1)
(√2 - √5)x + (√3 + √2)y = 0---(2)
Multiplying equation (1) by (√3 + √2) & equation (2) (√2 - √5), we get
(√3 + √2)2x - (√3 + √2) (√2 - √5)y = 0---(3)
(√2 - √5)2x + (√3 + √2) (√2 - √5)y = 0---(4)
By adding equation (3) and (4) we get
((√3 + √2)2 + (√2 - √5)2)x = 0
⇒ x = 0
(\(\because\) (√3 + √2)2 > 0 & (√2 - √5)2 > 0 ⇒ (√3 + √2)2 + (√2 - √5)2 \(\neq0\))
Put x = 0 in equation (2) we get
(√3 + √2)y = 0
⇒ y = 0 (\(\because\) \(\sqrt3+\sqrt2\neq0\))
Hence, solution of given system of equations is x = 0, y = 0