Question

What is the solution for the following pair of linear equations?

$\frac{1}{2x}-\frac{1}{y}=-1$

$\frac{1}{x}+\frac{1}{2y}=8$

(Where $x\ne 0,y\ne 0)$

• A. $x=\frac{1}{6}$ and $y=\frac{1}{4}$
• B. $x=\frac{1}{7}$ and $y=\frac{1}{2}$
• C. $x=\frac{1}{8}$ and $y=\frac{1}{3}$
• D. $x=\frac{3}{4}$ and $y=\frac{5}{4}$

Answer 1

The correct option is A $x=\frac{1}{6}$ and $y=\frac{1}{4}$

We have
$\frac{1}{2x}-\frac{1}{y}=-1\dots .\left(1\right)$

$\frac{1}{x}+\frac{1}{2y}=8\dots .\left(2\right)$

$\text{From equation (1),}\frac{1}{2x}=-1+\frac{1}{y}$
$\Rightarrow \frac{1}{2x}=\frac{(-y+1)}{y}$

$\Rightarrow 2x=\frac{y}{(-y+1)}$

$\Rightarrow x=\frac{y}{(-2y+2)}$

$\text{Putting,}x=\frac{y}{(-2y+2)}\text{in equation (2)}$,
$\Rightarrow \frac{(-2y+2)}{y}+\frac{1}{2y}=8$

$\Rightarrow \frac{(-4y+4+1)}{2y}=8$

$\Rightarrow -4y+4+1=16y$

$\Rightarrow 20y=5$

$\Rightarrow y=\frac{1}{4}$

$\text{Putting,}y=\frac{1}{4}\text{in equation (1),}$
$\Rightarrow \frac{1}{2x}-4=-1$
$\Rightarrow \frac{1}{2x}=3$
$\Rightarrow x=\frac{1}{6}$