Question

Solve the following equations:
$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{10}{3}$,
$x+y=10$.

• A. $x=2;y=5$
• B. $x=5;y=2$
• C. $x=9;y=1$
• D. $x=1;y=9$

Answer 1

Answer: (C), (D)

The correct options are
C $x=9;y=1$
D $x=1;y=9$

Let $x+y=10$ .......(i)

and $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{10}{3}$ .....(ii)

Put $\sqrt{\frac{x}{y}}=t$ and $\sqrt{\frac{y}{x}}=\frac{1}{t}$

$\Rightarrow t+\frac{1}{t}=\frac{10}{3}\Rightarrow \frac{t^2+1}{t}=\frac{10}{3}\Rightarrow 3t^2-10t+3=0\Rightarrow 3t^2-9t-t+3=0\Rightarrow 3t(t-3)-(t-3)=0\Rightarrow (3t-1)(t-3)=0\Rightarrow t=\frac{1}{3},3\Rightarrow \sqrt{\frac{x}{y}}=t\Rightarrow \frac{x}{y}=t^2$

For $t=\frac{1}{3}$, we have

$\frac{x}{y}=\frac{1}{9}\Rightarrow y=9x$ ..........(iii)

For $t=3$, we have

$\frac{x}{y}=9\Rightarrow x=9y$ ..........(iv)

Solving (i) and (iii), we have

$x=1$ and $y=9$

So, $(1,9)$ is one solution

Solving (i) and (iv), we get

$y=1$ and $x=9$

So, $(9,1)$ is the other solution.