Question

Root 2 x + root 3 y is equal to zero and root 3 x minus root 8 y is equal to zero solve

Answer 1

Answer:

The value of x and y is zero.                

Step-by-step explanation:

Given : Equations $\sqrt{2}x+\sqrt{3}y=0$ and $\sqrt{3}x-\sqrt{8}y=0$

To find : Solve the equations?

Solution :

$\sqrt{2}x+\sqrt{3}y=0$ ......(1)

$\sqrt{3}x-\sqrt{8}y=0$ ........(2)

Multiply equation (1) by $\sqrt{3}$ and equation (2) by $\sqrt{2}$

$\sqrt{6}x+\sqrt{9}y=0$ ......(3)

$\sqrt{6}x-\sqrt{16}y=0$ ........(4)

Subtract equation (3) and (4),

$\sqrt{6}x+\sqrt{9}y-\sqrt{6}x+\sqrt{16}y=0$

$3y+4y=0$

$7y=0$

$y=0$

Substitute in equation (1),

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

$\sqrt{2}x=0$

$x=0$

Therefore, The value of x and y is zero.

Answer 2

Given equation :

$\sqrt{2}x+\sqrt{3}y=0 .....(1)\\ \sqrt{3}x-\sqrt{8}y=0 .....(2 )$

to find : the value of x and y

solution :

on multiplying $\sqrt{2}$ in equation (1)  and  $\sqrt{3}$  in equation (2) and subtract  we get,

$\sqrt{6}x+\sqrt{9}y=0 .........(3)\\ \sqrt{6}x-\sqrt{16}y=0........(4)$

on subtracting equation (3) and equation (4)..

$= > 7y=0\\= > y=0$

on putting the value of y in equation (1) we get ,

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

Hence , the value of x , y is zero

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