Solve:$\frac{3 x}{2}-\frac{5 y}{3}=-2$$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Given: Pair of linear equations: 

$\frac{3 x}{2}-\frac{5 y}{3}=-2$

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

To do: To solve the given pair of linear equations.

Solution:

Given equations are:

$\frac{3 x}{2}-\frac{5 y}{3}=-2$  .......... $( i)$

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$  .......$( ii)$

On multiplying equation $( i)$ by $\frac{1}{3}$ 

$\frac{3 x}{2}\times\frac{1}{3}-\frac{5 y}{3}\times\frac{1}{3}=-2\times\frac{1}{3}$

$\frac{3}{6}x-\frac{5}{9}y=-\frac{2}{3}$  .......... $( iii)$

And multiply $( ii)$ by $\frac{3}{2}$

$\frac{x}{3}\times\frac{3}{2}+\frac{y}{2}\times\frac{3}{2}=\frac{13}{6}\times\frac{3}{2}$

$\frac{3}{6}x+\frac{3}{4}y=\frac{39}{12}$  ......... $(iv)$

Subtract $( iii)$ from $( iv)$

$\frac{3}{6}x+\frac{3}{4}y-\frac{3}{6}x+\frac{5}{9}y=\frac{39}{12}-( -\frac{2}{3})$

$\Rightarrow \frac{3}{4}y+\frac{5}{9}y=\frac{39}{12}+\frac{2}{3}$

$\Rightarrow \frac{27+20}{36}y=\frac{117+24}{36}$

$\Rightarrow \frac{47}{36}y=\frac{141}{36}$

$\Rightarrow 47y=141$

$\Rightarrow y=\frac{141}{47}$

$\Rightarrow y=3$

On putting $y=1$, in $( i)$

$\frac{3 x}{2}-\frac{5}{3}\times3=-2$

$\Rightarrow \frac{3 x}{2}-\frac{15}{3}=-2$

$\Rightarrow \frac{3x}{2}=-2+\frac{15}{3}$

$\Rightarrow \frac{3x}{2}=\frac{-6+15}{3}$

$\Rightarrow \frac{3x}{2}=\frac{9}{3}$

$\Rightarrow 9x=18$

$\Rightarrow x=\frac{18}{9}$

$\Rightarrow x=2$

Thus, $x=2,\ y=3$.