The vertices $A\left(x,2\right);B\left(3,4\right)\text{and}C\left(3,y\right),D\left(4,2\right)$   forms a parallelogram then $x+y$ is

$A\left(x,2\right);B\left(3,4\right)\text{and}C\left(3,y\right),D\left(4,2\right)$   forms a parallelogram then

Midpoint of  $\overline{AC}=\text{mid point of }\overline{BD}$  

$\left(\frac{x+3}{2},\frac{2+y}{2}\right)=\left(\frac{3+4}{2},\frac{4+2}{2}\right)$

$\left(\frac{x+3}{2},\frac{2+y}{2}\right)=\left(\frac{7}{2},3\right)$

$\frac{x+3}{2}=\frac{7}{2}$  and $\frac{2+y}{2}=3$

$x+3=7\text{and}2+y=6$

$x=4\text{and}y=4$

$\therefore x+y=4+4=8$