Question

solve the following pair of linear equations by the substitution method.
$\sqrt{2 \: } x \: + \: \sqrt{3} \: y \: = \: 0$
$\sqrt{3} x \: - \: \sqrt{8} y \: = \: 0$
​

Answer 1

$\large \green { \fcolorbox{purple}{WHITE}{ \: \textsf{\ \ \ \ddag \ \ \ \ \ \red{QUESTION}\ \ \ \ \ \ddag \ \ \ }}}$

☞Solve the following pair of linear equations by the substitution method.

$\sqrt{2 \: } x \: + \: \sqrt{3} \: y \: = \: 0$

$\sqrt{3} x \: - \: \sqrt{8} y \: = \: 0$.

____________________________________

$\large \green { \fcolorbox{purple}{WHITE}{ \: \textsf{\ \ \ \ddag \ \ \ \ \ \purple{SOLUTION}\ \ \ \ \ \ddag \ \ \ }}}$

⟼$\sqrt{2 \: } x \: + \: \sqrt{3} \: y \: = \: 0$___(1)

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

⟼From equation (1), we obtain

$⇒x= \frac{-\sqrt{3y}}{\sqrt{2}}$

⟼Substituting this value in equation (2), we obtain

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

$⇒\frac{\sqrt{3y}}{\sqrt{2}}- 2\sqrt{2y}=0$

$⇒y(-\frac{3}{\sqrt{2}}-2\sqrt{2})=0$

$⇒y=0$

⟼Substituting this value in equation (3), we obtain $x=0$

⟹Therefore, $x=0\: and\: y=0$

_________________________________

Hope it helps you ☺️!

Answer 2

Step-by-step explanation:

Answer is in the attachment. Rude words are coming so