Answer
$(8, -2)$
Work Step by Step
Given system is-
$\frac{3}{4}x +\frac{1}{2}y$ = $5$ __ eq.1
$-\frac{1}{4}x -\frac{3}{2}y$ = $1$ __ eq.2
Multiplying eq.1 by '3'-
$\frac{9}{4}x +\frac{3}{2}y$ = $15$ __ eq.3
Adding eq.2 and eq.3 -
$-\frac{1}{4}x -\frac{3}{2}y+\frac{9}{4}x +\frac{3}{2}y$ = $1+15$
i.e $-\frac{1}{4}x +\frac{9}{4}x $ = $16$
i.e.$x(-\frac{1}{4} +\frac{9}{4}) $ = $16$
i.e.$x(\frac{-1+9}{4} ) $ = $16$
i.e.$\frac{8}{4}x $ = $16$
i.e. $2x$ = $16$
i.e. $x$ = $8$
Substituting for $x$ in eq.1
$\frac{3}{4}(8) +\frac{1}{2}y$ = $5$
i.e. $\frac{24}{4} +\frac{1}{2}y$ = $5$
i.e. $6+\frac{1}{2}y$ = $5$
i.e. $\frac{1}{2}y$ = $5-6$= $-1$
i.e. $y$ = $-1\times2$ = $-2$
Thus $(8, -2)$ is the solution of given system.
Alternate Method-
Given system is-
$\frac{3}{4}x +\frac{1}{2}y$ = $5$ __ eq.1
$-\frac{1}{4}x -\frac{3}{2}y$ = $1$ __ eq.2
Multiplying eq.1 by 'LCM' of 4 and 2 i.e. '4'-
$3x +2y$ = $20$ __ eq.3
Multiplying eq.2 by 'LCM' of 4 and 2 i.e. '4'-
$-x -6y$ = $4$ __ eq.4
Multiplying eq.4 by '3'-
$-3x -18y$ = $12$ __ eq.5
Adding eq.3 and eq.5-
$ -16y$ = $32$
i.e. $ y$ = $-2$
Substituting for y in eq.3-
$3x +2(-2)$ = $20$
i.e. $3x -4$ = $20$
i.e. $3x $ = $24$
i.e. $x $ = $8$
Thus $(8, -2)$ is the solution of given system.