Question

$$\begin{gathered} 2x-\frac{3}{y}=9, \\ 3x+\frac{7}{y}=2,y\ne 0. \end{gathered}$$

Answer 1

The given equations are:
$2x-\frac{3}{y}=9$ ............(i)
$3x+\frac{7}{y}=2$ .............(ii)
Putting $\frac{1}{y}=v$, we get:
2x − 3v = 9 .............(iii)
3x + 7v = 2 ...........(iv)

On multiplying (iii) by 7 and (iv) by 3, we get:
14x − 21v = 63 .............(v)
9x + 21v = 6............(vi)

On adding (v) and (vi), we get:
23x = 69 ⇒ x = 3

On substituting x = 3 in (i), we get:
$2\times 3-\frac{3}{y}=9$
$\Rightarrow 6-\frac{3}{y}=9\Rightarrow \frac{3}{y}=-3\Rightarrow y=-1$
Hence, the required solution is x = 3 and y = −1.