Question

Find the values of $x$ and $y$, if

$\frac{2}{x}+\frac{6}{y}=13;\frac{3}{x}+\frac{4}{y}=12$

• A. $x=\frac{1}{6},y=\frac{2}{3}$
• B. $x=\frac{1}{3},y=\frac{2}{3}$
• C. $x=\frac{1}{7},y=\frac{2}{3}$
• D. $x=\frac{1}{2},y=\frac{2}{3}$

Answer 1

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

$\frac{2}{x}+\frac{6}{y}=13$

$\frac{3}{x}+\frac{4}{y}=12$

Lets assume, $\frac{1}{x}=a,\frac{1}{y}=b$, then the above equations reduces to

$2a+6b=13$ ...(1)

$3a+4b=12$ ...(2)

Multiplying (1) by $3$ and (2) by $2$, we get

$6a+18b=39$ ...(3)

$6a+8b=24$ ...(4)

Subtracting (3) and (4), we get

$10b=15\Rightarrow b=\frac{3}{2}$

Now, $2a=13-6\times \frac{3}{2}=13-9=4$

$\Rightarrow a=2$

So, $x=\frac{1}{2};y=\frac{2}{3}$

Hence, option D is the correct answer.