Question

solve the pair of equation...
$\frac{2}{ \sqrt{x} } + \frac{3}{\sqrt{y} } = 2 \\ \\ \frac{4}{ \sqrt{x} } - \frac{9}{ \sqrt{y} } = - 1$
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Answer 1

Simplify:

Step-by-step explanation:

$\ Given \ value:$

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

$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1........(ii)\\\\\ find: \\\\\ x \ = \ ? \\\\ \ y \ = \ ?$

$\ Solution: \\\\\ multiply \ by \ 3 \ in \ the \ equation \ (i)...\\\\3 \times (\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}})=3 (2)\\\\\frac{6}{\sqrt{x}}+\frac{9}{\sqrt{y}}=6..................(iii)\\\\\ Subtract \ equation \ (iii) \ to \ (ii)$

$\frac{6}{\sqrt{x}}+\frac{9}{\sqrt{y}}=6 \\\\\ - \\\\\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}= -1 \\\\\frac{2}{\sqrt{x}}=6\\\\\sqrt{x}=3$

$\sqrt{x}=3\\\\\ put \ the \ value \ of \sqrt{x} \ in \ equation \ (i) \\\\ \frac{2}{3}+\frac{3}{\sqrt{y}} =2 \\\\\frac{3}{\sqrt{y}} =2 - \frac{2}{3} \\\\\frac{3}{\sqrt{y}} = \frac{6-2}{3} \\\\\frac{3}{\sqrt{y}} = \frac{4}{3} \\\\\sqrt{y}= \frac{4}{9}$

$\ x \ = 9 \ and \ y \ = \frac{16}{81}$

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