Question

root2x+root3y=0
root3x- root8y=0
solve it in substitution method

Answer 1

Yehi hoga bhai thora fast tha isliye writing garvar hai

Answer 2

x = 0 and y = 0

Step-by-step explanation:

The given equations are:

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

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

To find, the values of x and y = ?

∴ $\sqrt{2}x+\sqrt{3}y=0$

⇒ $\sqrt{2}x=-\sqrt{3}y$

⇒ $x=\dfrac{-\sqrt{3}y}{\sqrt{2}}$

Putting the value of x in equation (2), we get

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

⇒ $\dfrac{-3y-16y}{\sqrt{2}}=0$

⇒ 19y = 0

⇒ y = 0

Put y = 0 in equatiion (1), we get

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

⇒ $\sqrt{2}x+0=0$

⇒ x = 0

∴ x = 0 and y = 0