Solve this

Given: $3 x+a y=4$

$2 x+a y=2$

Using Cramer's rule, we get

$\mathrm{D}=\mid 3 \mathrm{a}$

$2 \mathrm{a} \mid=3 \mathrm{a}-2 \mathrm{a}=\mathrm{a}$

$\mathrm{D}_{1}=\mid 4 \mathrm{a}$

$2 \mathrm{a} \mid=4 \mathrm{a}-2 \mathrm{a}=2 \mathrm{a}$

$\mathrm{D}_{2}=\mid 3 \mathrm{4}$

$2 \quad 2 \mid=6-8=-2$

Now,

$\mathrm{x}=\frac{\mathrm{D}_{1}}{\mathrm{D}}=\frac{2 \mathrm{a}}{\mathrm{a}}=2$

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

$\therefore \mathrm{x}=2$ and $\mathrm{y}=-\frac{2}{\mathrm{a}}$