Question

Solve the following pair of equations.
$\frac{1}{x}+\frac{3}{y}=1$
$\frac{6}{x}-\frac{12}{y}=2$
(where $x\ne 0,y\ne 0)$

• A. $x=\frac{3}{5},y=\frac{7}{3}$
• B. $x=\frac{5}{3},y=\frac{15}{2}$
• C. $x=3,y=11$
• D. $x=4,y=9$

Answer 1

The correct option is B

$x=\frac{5}{3},y=\frac{15}{2}$

Let $\frac{1}{x}=a$ and $\frac{1}{y}=b$ (As $x\ne 0,y\ne 0)$

Then, the given equations become

$a+3b=1$ …(1)

$6a-12b=2$ …(2)

On multiplying equation (1) by 4, we get

$4a+12b=4$ …(3)

On adding equation (2) and equation (3), we get

$10a=6$

$\Rightarrow a=\frac{3}{5}$

On substituting $a=\frac{3}{5}$ in equation (1), we get

$\frac{3}{5}+3b=1$

$\Rightarrow b=\frac{1-\left(\frac{3}{5}\right)}{3}$

$\Rightarrow b=\frac{2}{15}$

We know that $\frac{1}{x}=a$ and $\frac{1}{y}=b$.

Hence, $x=\frac{5}{3}$ and $y=\frac{15}{2}$.