Solve this

Given : $9 x+5 y=10$

$3 y-2 x=8$

Rearranging the second equation, the two equations can be written as

$9 x+5 y=10$

$-2 x+3 y=8$

Now,

$D=\mid 9 \quad 5$

$-2 \quad 3 \mid=27+10=37$

$D_{1}=\mid 105$

$8 \quad 3 \mid=30-40=-10$

$D_{2}=\mid 9 \quad 10$

$-2 \quad 8 \mid=72+20=92$

Using Cramer's rule, we get

$x=\frac{D_{1}}{D}=\frac{-10}{37}$

$y=\frac{D_{2}}{D}=\frac{92}{37}$

$\therefore x=\frac{-10}{37}$ and $y=\frac{92}{37}$