Answer
$(2, 1)$
Work Step by Step
Given system is-
$3x +2y$ = $8$ __ eq.1
$x -2y$ = $0$ __ eq.2
Adding eq.1 and eq.2 -
$3x +2y+x -2y$ = $8+0$
i.e. $4x$ = $8$
i.e. $x$ = $\frac{8}{4}$
i.e. $x$ = $2$
Substituting for $x$ in eq.1
$3(2) +2y$ = $8$
i.e. $6 +2y$ = $8$
i.e. $2y$ = $8-6$ =$2$
i.e. $y$ = $\frac{2}{2}$
i.e. $y$ = $1$
Thus $(2, 1)$ is the solution of given system.