Question

Solve the following pair of linear equation:
$\frac{3}{\sqrt{x}}+\frac{4}{\sqrt{y}}=2,\frac{5}{\sqrt{x}}+\frac{7}{\sqrt{y}}=\frac{41}{12},x>0,y>0$

• A. $x=9,y=16$
• B. $\sqrt{x}=3,\sqrt{y}=4$
• C. $x=2,y=5$
• D. $x=4,y=25$

Answer 1

Answer: (A), (B)

The correct options are
A $x=9,y=16$
B $\sqrt{x}=3,\sqrt{y}=4$

Given equations are:

$4\sqrt{x}+3\sqrt{y}=2\sqrt{x}\sqrt{y}$...(1)

$84\sqrt{x}+60\sqrt{y}=41\sqrt{x}\sqrt{y}$...(2)

$\left(1\right)\times 21\left(2\right)$

$84\sqrt{x}+63\sqrt{y}=42\sqrt{x}\sqrt{y}$

$84\sqrt{x}+60\sqrt{y}=41\sqrt{x}\sqrt{y}$

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$3\sqrt{y}=\sqrt{x}\sqrt{y}$

$\Rightarrow \sqrt{x}=3$

$\Rightarrow x=9$

Substituting the value of $\sqrt{x}$ in equation (1), we get,

$4\sqrt{x}+3\sqrt{y}=2\sqrt{x}\sqrt{y}$

$4\times 3+3\sqrt{y}=2\times 3\sqrt{y}$

$12+3\sqrt{y}=6\sqrt{y}$

$\Rightarrow \sqrt{y}=4$

$\Rightarrow y=16$