Solve this

Given: $2 x-y=-2$

$3 x+4 y=3$

Using Cramer's Rule, we get

$\mathrm{D}=\mid 2-1$

$3 \quad 4 \mid=8+3=11$

$\mathrm{D}_{1}=\mid-2-1$

$3 \quad 4 \mid=-8+3=-5$

$\mathrm{D}_{2}=\mid 2-2$

$3 \quad 3 \mid=6+6=12$

Now,

$\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{-5}{11}$

$\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{12}{11}$

$\therefore \mathrm{x}=-\frac{5}{11}$ and $\mathrm{y}=\frac{12}{11}$