Answer
$(12, 4)$
Work Step by Step
Given system is-
$-4x +12y$ = $0$ __ eq.1
$12x +4y$ = $160$ __ eq.2
Multiplying eq.1 by '3'-
$-12x +36y$ = $0$ __ eq.3
Adding eq.2 and eq.3 -
$12x +4y-12x +36y$ = $160+0$
i.e. $40y$ = $160$
i.e. $y$ = $\frac{160}{40}$
i.e. $y$ = $4$
Substituting for $y$ in eq.1
$-4x +12(4)$ = $0$
i.e. $-4x $ = $-48$
i.e. $4x$ = $48$
i.e. $x$ = $\frac{48}{4}$
i.e. $x$ = $12$
Thus $(12, 4)$ is the solution of given system.