Question

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

Answer 1

For equations
$\frac{3}{\sqrt{x}}+\frac{4}{\sqrt{y}}=\mathrm{2.....}\left(i\right)$
$\frac{5}{\sqrt{x}}+\frac{7}{\sqrt{y}}=\frac{41}{12}...\left(ii\right)(x>0,y>0)$
Multiplying $\left(i\right)$ by $5$ and $\left(ii\right)$ by $3$
$\frac{15}{\sqrt{x}}+\frac{20}{\sqrt{y}}=\mathrm{10...}\left(i\right)$
$\frac{15}{\sqrt{x}}+\frac{21}{\sqrt{y}}=\frac{41}{4}.....\left(ii\right)$
$\left(ii\right)$- $\left(i\right)$ subtracting the eq(i) from eq(ii) we get,
$\frac{1}{\sqrt{y}}=\frac{1}{4}\Rightarrow y=16$
Substituting the value of $y$ in $\left(i\right)$, we get
$x=9$

Hence, the values of $x$ and $y$ are $16$ and $9$ respectively.