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

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