Question

solve by the method of elimination & comparison:

(√3+√2)x-(√2-√5)y=0
(√2-√5)x+(√3+√2)y=0​

Answer 1

Answer:

x=0,y=0

Step-by-step explanation:

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Answer 2

Answer:

The correct answer to this question is $x = 0, y = 0$

Step-by-step explanation:

Given - $(\sqrt{3} +\sqrt{2} ) x- (\sqrt{2} - \sqrt{5} )y = 0$ ----- $1)$

$(\sqrt{2} - \sqrt{5} ) x+ (\sqrt{3} - \sqrt{2} )y = 0$ ------$2)$

Multiplying equation $(1)$ by $(\sqrt{3} +\sqrt{2} )$ & equation $(2)$ $(\sqrt{2} - \sqrt{5} )$, we get

$(\sqrt{3} +\sqrt{2} )^{2}$$x$ - $(\sqrt{3} +\sqrt{2} )$ $(\sqrt{2} - \sqrt{5} )$$y$  = $0 ---(3)$

$(\sqrt{2} - \sqrt{5} )^{2} x$ + $(\sqrt{3} +\sqrt{2} )$ $(\sqrt{2} - \sqrt{5} )$$y = 0 ---(4)$

By adding equation $3$ and $4$

$(\sqrt{3} +\sqrt{2} )^{2}$ $+$ $(\sqrt{2} - \sqrt{5} )^{2}$$x = 0$

( ∵  $(\sqrt{3} +\sqrt{2} )$$2$ > 0 &  $(\sqrt{2} - \sqrt{5} )$$2$ > 0 ⇒ $(\sqrt{3} +\sqrt{2} )$$2$ + $(\sqrt{2} - \sqrt{5} )$$2$  ≠ 0 ≠0)

Put x = 0 in equation (2) we get

$(\sqrt{3} +\sqrt{2} )y$ $= 0$

⇒ $y = 0$ ( ∵ ∵  $(\sqrt{3} +\sqrt{2} )$ ≠ 0 $3+2$≠0)

Hence, solution of given system of equations is $x = 0, y = 0$

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