Solve the following systems of equations:

The given equations are:

$99 x+101 y=499 \ldots(i)$

$101 x+99 y=501 \ldots(i i)$

Multiply equation $(i)$ by 99 and equation $(i i)$ by 101 , and subtract (ii) from (i) we get

Put the value of $x$ in equation $(i)$, we get

$99 \times 3+101 y=499$

$\Rightarrow 101 y=202$

$\Rightarrow y=2$

Hence the value of $x=3$ and $y=2$