Question:
Solve the following systems of equations:
$2 x-\frac{3}{y}=9$
$3 x+\frac{7}{y}=2, y \neq 0$
Solution:
The given equations are:
$2 x-\frac{3}{y}=9 \ldots(i)$
$3 x+\frac{7}{y}=2 \ldots(i i)$
Multiply equation $(i)$ by 3 and $(i i)$ by 2 and subtract equation (ii) from (i) we get
$6 x-\frac{9}{y}=27$
Put the value of $y$ in equation $(i)$, we get
$2 x-\frac{3}{-1}=9$
$\Rightarrow 2 x=6$
$\Rightarrow x=3$
Hence the value of $x=3$ and $y=-1$
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