Solve by the method of elimination : (√3 + √2)x – (√2 - √5

1.	Solve by the method of elimination : (√3 + √2)x – (√2 - √5 ) y = 0; (√2 - √5 )x + (√3 + √2) y = 0
Answer» (√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

Solve by the method of elimination : (√3 + √2)x – (√2 - √5 ) y = 0; (√2 - √5 )x + (√3 + √2) y = 0

Answer»

(√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

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