Question

Solve the pair of equations by reducing them to a pair of linear equations.
2/√x + 3/√y = 2
4/√x - 9/√y = -1

Answer 1

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$\frac{2}{ \sqrt{x} } + \frac{3}{ \sqrt{y} } = 2 \\ \frac{4}{ \sqrt{x} } - \frac{9}{ \sqrt{y} } = - 1 \\ \\ first \ \: \\ let \: \frac{1}{ \sqrt{x} } = u \: \: and \: \frac{1}{ \sqrt{y} } = v \\ Now ,\: the \: equations \: becomes : \\ \\ = > 2u + 3v = 2 - - > (1) \\ = > 4u - 9v = - 1 - - > (2)\\ \\ USING \: SUBSTITUTION\: METHOD : \\ \\ From \: eq.(1) \\ u = \frac{2 - 3v}{2} - - > (3) \\ \\ Putting, \: the \: value \: of \:' u' \: in \: eq.(2) \\ \\ = > 4u - 9v = - 1 \\ = > 4( \frac{2 - 3v}{2} ) - 9v = - 1 \\ \\ = > 8 - 12v - 18v = - 2 \\ = > - 30v = - 10 \\ = > v = \frac{1}{3} \\ \\ Putting \: the \: value \: of \: 'v' \: in \: eq.(3) \\ = > u = \frac{2 - 3( \frac{1}{3}) }{2} \\ = > u = \frac{1}{2} \\ \\ so \: u \: = \frac{1}{2} and \: v = \frac{1}{3} \\ \\ Now \: \frac{1}{ \sqrt{x} } = u \\ = > \frac{1 }{ \sqrt{x} } = \frac{1}{2} \\ = > \sqrt{x} = 2 \\ = > x = 4 \\ \\ and \: \frac{1}{ \sqrt{y} } = v \\ = > \frac{1}{ \sqrt{y} } = \frac{1}{3} \\ = > \sqrt{y} = 3 \\ = > y = 9$

So ,x = 4 and y = 9 ...........
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