Question

$$\begin{gathered} \frac{5}{x}+6y=13, \\ \frac{3}{x}+4y=7,x\ne 0. \end{gathered}$$

Answer 1

The given equations are:
$\frac{5}{x}+6y=13$ ............(i)
$\frac{3}{x}+4y=7$ .............(ii)

Putting $\frac{1}{x}=u$, we get:
5u + 6y = 13 .............(iii)
3u + 4y = 7 ...........(iv)

On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ...........(v)
18u + 24y = 42 ............(vi)

On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
$\Rightarrow \frac{1}{x}=5\Rightarrow x=\frac{1}{5}$
On substituting $x=\frac{1}{5}$ in (i), we get:
$\frac{5}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}+6y=13$
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
⇒ y = −2

Hence, the required solution is $x=\frac{1}{5}$ and y = −2.