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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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