Question

Solution for the following pairs of equations is.
$\frac{1}{x}+\frac{3}{y}=1$
$\frac{6}{x}-\frac{12}{y}=2$
(Where $x\ne 0,y\ne 0)$

• A. $x$ =$\frac{3}{5}$,$y$ =$\frac{7}{3}$
• B. $x$ =$\frac{5}{3}$,$y$ =$\frac{7}{2}$
• C. $x$ =$\frac{5}{3}$,$y$ =$\frac{15}{2}$
• D. $x$ =$\frac{5}{3}$,$y$ =$\frac{12}{2}$

Answer 1

The correct option is C $x$ =$\frac{5}{3}$,$y$ =$\frac{15}{2}$

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)

Multiplying equation (1) by 4, we get $4a+12b=4$ …(3)
Adding equation (2) and equation (3), we get $10a=6$
$\Rightarrow a=\frac{3}{5}$

Putting $a=\frac{3}{5}$ in the equation (1), we get
$\frac{3}{5}+3b=1$
$\Rightarrow b=\frac{1-\left(\frac{3}{5}\right)}{3}$

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

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