Question

$$\begin{gathered} \frac{1}{7x}+\frac{1}{6y}=3, \\ \frac{1}{2x}-\frac{1}{3y}=5\left(x\ne 0,y\ne 0\right). \end{gathered}$$

Answer 1

The given equations are:
$\frac{1}{7x}+\frac{1}{6y}=3$............(i)
$\frac{1}{2x}-\frac{1}{3y}=5$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
$\frac{u}{7}+\frac{v}{6}=3$.............(iii)
On multiplying (iii) by 42, we get:
6u + 7v = 126 .........(iv)

Again, $\frac{u}{2}-\frac{v}{3}=5$ .............(v)

On multiplying (v) by 6, we get:
3u − 2v = 30 .............(vi)

On multiplying (iv) by 2 and (vi) by 7, we get:
12u + 14v = 252 ..............(vii)
21u − 14v = 210 ................(viii)

On adding (vii) and (viii), we get:
33u = 462 ⇒ u = 14
$\Rightarrow \frac{1}{x}=14\Rightarrow x=\frac{1}{14}$

On substituting $x=\frac{1}{14}$ in (i), we get:
$\frac{1}{7\times {\displaystyle \frac{1}{14}}}+\frac{1}{6y}=3$
⇒ $2+\frac{1}{6y}=3\Rightarrow \frac{1}{6y}=\left(3-2\right)=1$
$\Rightarrow y=\frac{1}{6}$

Hence, the required solution is $x=\frac{1}{14}$ and $y=\frac{1}{6}$.