Solve this

Given : $2 x+3 y=10$

$x+6 y=4$

Using Cramer's Rule, we get

$\mathrm{D}=\mid 2 \quad 3$

$1 \quad 6 \mid=12-3=9$

$\mathrm{D}_{1}=\mid 10 \quad 3$

$4 \quad 6 \mid=60-12=48$

$\mathrm{D}_{2}=\mid 2 \quad 10$

$1 \quad 4 \mid=8-10=-2$

Now,

$\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{48}{9}=\frac{16}{3}$

$\mathrm{y}=\frac{\mathrm{D}_{2}}{\mathrm{D}}=\frac{-2}{9}$

$\therefore x=\frac{16}{3}$ and $\mathrm{y}=\frac{-2}{9}$