Question

$$\begin{gathered} \frac{4}{x}+3y=8, \\ \frac{6}{x}-4y=-5. \end{gathered}$$

Answer 1

The given equations are:
$\frac{4}{x}+3y=8$ ............(i)
$\frac{6}{x}-4y=-5$ .............(ii)
Putting $\frac{1}{x}=u$, we get:
4u + 3y = 8 .............(iii)
6u − 4y = −5 ...........(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
16u + 12y = 32 ...........(v)
18u − 12y = −15 ............(vi)

On adding (v) and (vi), we get:
34u = 17
$\Rightarrow u=\frac{17}{34}=\frac{1}{2}$
$\Rightarrow \frac{1}{x}=\frac{1}{2}\Rightarrow x=2$

On substituting x = 2 in (i), we get:
$\frac{4}{2}+3y=8$
⇒ 2 + 3y = 8
⇒ 3y = (8 − 2) = 6
⇒ y = 2
Hence, the required solution is x = 2 and y = 2.