[tex] \qquad \qquad \huge \underline{\boxed{ \sf {Answer}}}[/tex]
Let's solve ~
The given equations are :
[tex] \qquad \dashrightarrow \bf \: 3x + 4y = 36 \sf \: \: \: \: \: \: \: \: \: \: (1)[/tex]
and
[tex] \qquad \dashrightarrow \sf \:y = - \frac{1}{2} x + 8[/tex]
[ multiply each sides by 2 ]
[tex] \qquad \dashrightarrow \sf \: 2y = - x + 16[/tex]
[tex] \qquad \dashrightarrow \sf \:x + 2y = 16[/tex]
[ multiply both sides by 2 ]
[tex] \qquad \dashrightarrow \bf \: 2x + 4y = 32 \: \: \: \: \: \: \: \: \: \: \: \sf(2)[/tex]
Now, subtract equation (2) from (1)
[tex] \qquad \dashrightarrow \sf \: 3x + 4y - (2x + 4y) = 36 - 32[/tex]
[tex] \qquad \dashrightarrow \sf \: 3x + \cancel{4y }- 2x - \cancel{4y }= 4[/tex]
[tex] \qquad \dashrightarrow \bf \: x = 4[/tex]
Now, as we got the value of x, let's solve for y :
[tex] \qquad \dashrightarrow \sf \: y = - \dfrac{1}{2} x+ 8[/tex]
[tex] \qquad \dashrightarrow \sf \: y = -\dfrac{1}{2} (4)+8[/tex]
[tex] \qquad \dashrightarrow \sf \: y = - 2 + 8[/tex]
[tex] \qquad \dashrightarrow \bf \: y = 6[/tex]
