Question

Solve the following pair of linear equations:
$\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$
$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$
(Where $x>0,y>0)$

• A. $x=9,y=4$
• B. $x=4,y=9$
• C. $x=3,y=2$
• D. $x=\frac{1}{2},y=\frac{1}{3}$

Answer 1

The correct option is B $x=4,y=9$

Given,
$\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$
$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$
(Where $x>0,y>0)$

The pair of equations are not linear. We will substitute
$\frac{1}{x}$ as $u^2$ and $\frac{1}{y}$ as $v^2$ (As $x>0,y>0)$
Then we will get the equations as
$2u+3v=2$.........(i)
$4u-9v=-1$.........(ii)

We will use method of elimination to solve the equation
Multiply equation (i) by 3
$6u+9v=6$.........(iii)

Adding (ii) and (iii) we get
$10u=5$
$\Rightarrow u=\frac{1}{2}$
Substituting u in (i), we get
$2\left(\frac{1}{2}\right)+3v=2$
$\Rightarrow 3v=1$
$\Rightarrow v=\frac{1}{3}$
So $x=\frac{1}{u^2}=4$
$y=\frac{1}{v^2}=9$