Question

Find the value of x, y in the pair of equation given below, where $x\ne 0\&y\ne 0$.
$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$

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

Answer 1

The correct option is A $x=\frac{1}{2},y=\frac{1}{3}$

Let $\frac{1}{x}$ be 'a' and $\frac{1}{y}$ be 'b' (As $x$ $\ne$ 0 & $y$ $\ne$ 0).

We will get the equations as,

$2a+3b=13$ ....(1)

$5a-4b=-2$ ....(2)

Multiplying the first equation by 4 and the second equation by 3, we get

$8a+12b=52$

$15a-12b=-6$

Now adding the two equations, we get

$23a=46$

$a=2$

Substituting $a=2$ in equation (1), we get,

$2\times 2+3b=13$

$3b=9$

$\Rightarrow b=3$

Earlier, we have assumed that $a=\frac{1}{x}$ and $b=\frac{1}{y}$

$\therefore$ $x=\frac{1}{2}$ and $y=\frac{1}{3}$