Question

Solve the following pair of equations by reducing them to a pair of linear equations:

$\frac{1}{2x}+\frac{1}{3y}=2,\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

• A. $x=\frac{1}{4}$ and $y=\frac{1}{7}$
• B. $x=\frac{2}{5}$ and $y=\frac{1}{3}$
• C. $x=\frac{3}{7}$ and $y=\frac{1}{7}$
• D. $x=\frac{1}{2}$ and $y=\frac{1}{3}$

Answer 1

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

Given pair of equations,

$\frac{1}{2x}+\frac{1}{3y}=2,\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$

We get ,

$\frac{1}{2}u+\frac{1}{3}v=23u+2v=12........\left(1\right)$

$\frac{1}{3}u+\frac{1}{2}v=\frac{13}{6}2u+3v=13.........\left(2\right)$

Multiplying $\left(1\right)$ by $3$ and $\left(2\right)$ by $2$,

$3(3u+2v)=3\times 12.........\left(3\right)$

$2(2u+3v)=2\times 13.........\left(4\right)$

then subtract $\left(4\right)$ from $\left(3\right)$, we get

$3(3u+2v)-2(2u+3v)=3\times 12-2\times 13$

$9u-4u=36-265u=10\Rightarrow u=2$

Then substituting $u=2$ in $\left(1\right)$, we get

$\Rightarrow 6+2v=12$

$\Rightarrow v=3$

Now, $u=2$ and $v=3$

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