Given,

$\surd 2\text{ }x\text{ }+\text{ }\surd 3\text{ }y\text{ }=\text{ }0\text{ }and\text{ }\surd 3\text{ }x\text{ }\text{ }\surd 8\text{ }y\text{ }=\text{ }0$

are the two equations.

From 1^st equation, we get,

$x\text{ }=\text{ }\text{ }\left( \surd 3/\surd 2 \right)y\text{ }\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left( 1 \right)$

Putting the value of x in the given second equation to get,

$\surd 3\left( -\surd 3/\surd 2 \right)y\text{ }\text{ }\surd 8y\text{ }=\text{ }0\Rightarrow \left( -3/\surd 2 \right)y-\text{ }\surd 8\text{ }y\text{ }=\text{ }0$

$\Rightarrow y\text{ }=\text{ }0$

Now, substitute the value of y in equation (1), we get,

$x\text{ }=\text{ }0$

Therefore, x = 0 and y = 0.

(vi) Given,

$\left( 3x/2 \right)-\left( 5y/3 \right)\text{ }=\text{ }-2\text{ }and\text{ }\left( x/3 \right)\text{ }+\text{ }\left( y/2 \right)\text{ }=\text{ }13/6$  are the two equations.

From 1^st equation, we get,

$\left( 3/2 \right)x\text{ }=\text{ }-2\text{ }+\text{ }\left( 5y/3 \right)$

$\Rightarrow x\text{ }=\text{ }2\left( -6+5y \right)/9\text{ }=\text{ }\left( -12+10y \right)/9\text{ }\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( 1 \right)$

Putting the value of x in the given second equation to get,

$\left( \left( -12+10y \right)/9 \right)/3\text{ }+\text{ }y/2\text{ }=\text{ }13/6$

$\Rightarrow y/2\text{ }=\text{ }13/6\text{ }\left( \text{ }\left( -12+10y \right)/27\text{ } \right)\text{ }+\text{ }y/2\text{ }=\text{ }13/6$

Now, substitute the value of y in equation (1), we get,

$\left( 3x/2 \right)\text{ }\text{ }5\left( 3 \right)/3\text{ }=\text{ }-2$

$\Rightarrow \left( 3x/2 \right)\text{ }\text{ }5\text{ }=\text{ }-2$

$\Rightarrow x\text{ }=\text{ }2$

Therefore, x = 2 and y = 3.