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# Solve the following pair of linear equations by the substitution method. root of 2 x + root of 3 y = 0 root of 3 x - root of 8 y = 0

Q1.    Solve the following pair of linear equations by the substitution method.

(v)    $\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

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Given, two equations,

$\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Now, from (1), we have

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

Substituting this in (2), we get

$\sqrt{3}x-\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )=0$

$\Rightarrow \sqrt{3}x=\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )$

$\Rightarrow \3x=-4x$

$\Rightarrow7x=0$

$\Rightarrow x=0$

Substituting this value of x in (3)

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

Hence, Solution of the given equations is,

$x=0,\:and \:y=0$.

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