Solve the pair of linear equations x + 2y = $\frac{3}{2}$ and 2x + y = 3 using the substitution method. 

The first equation is $\mathrm{x}+2\mathrm{y}=\frac{3}{2}$. 

⇒ 2(x+ 2y) = 3 

⇒ 2x+ 4y= 3 ... (1) 

The second equation is 2x+ y= 3. 

⇒ y= 3 − 2x... (2) 

On substituting equation (2) in (1), we get 

2x+ 4y = 3 

⇒ 2x+ 4(3 − 2x) = 3 

⇒ 2x+ 12 − 8x= 3 

⇒ 12 − 6x= 3 

⇒ −6x = 3 − 12 

⇒ −6x = −9 

$\Rightarrow \mathrm{x}=\frac{3}{2}$ 

Now using $\mathrm{x}=\frac{3}{2}$ in equation (2), we get 

$\mathrm{y}=3-2\times \frac{3}{2}=0$ 

Hence, $x=\frac{3}{2}$ and y = 0 is the required solution of the given equations.