Question

Solve the eqn ;-

$\sqrt{2}x + \sqrt{3} y = 0 \\ \\ \sqrt{3} x - \sqrt{8} y = 0$
Note :- Find (x) first .

Answer 1

Hey!

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Refer the pic for answer.

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Hope it helps...!!!

Answer 2

$\boxed{ \mathfrak{Hello \: Sister}}$

• Given that :-
$\sqrt{2x} + \sqrt{3} y = 0 \\ \\ \Rightarrow \: x = - \frac{ \sqrt{3} y}{ \sqrt{2} } \: ....... \: (1)$
putting this in Equation ( 2 )

That's -

√3x - √8y = 0

we obtained -

$\sqrt{3} \times \frac{ - \sqrt{3y} }{ \sqrt{2} } - \sqrt{8}y = 0 \\ \\ \Rightarrow \frac{ {( - 3)}^{2} y }{ \sqrt{2} } - \sqrt{8} y = 0 \\ \\ \Rightarrow \frac{3y - \sqrt{2} \times \sqrt{8} y }{ \sqrt{2} } = 0 \\ \\ \Rightarrow \: 3y - \sqrt{2} \times \sqrt{2} \times 2 \times y= 0 \\ \\ \Rightarrow 3y - 4y = 0 \\ \\ \Rightarrow - y = 0 \\ \\ \Rightarrow \boxed{ \bf y = 0}.....(3)$


Putting this value in equation ( 1 )

$x = \frac{ \sqrt{3}y }{ \sqrt{2} } \\ \\ \Rightarrow x = \sqrt{3} \times 0 \div \sqrt{2} \\ \\ \Rightarrow \boxed { \bf x = 0}$
$\boxed{ \bf Hence \: the \: value \: of \: x = \: 0 \: and y = 0} \\ \\ \mathfrak{ good \: luck : )}$