[tex]\boxed{y}=\cfrac{1}{2}x-4\\\\ y=-2x+1 \end{cases}\qquad \qquad \qquad \stackrel{\textit{substituting on the 2nd equation}}{\stackrel{\boxed{y}}{\cfrac{1}{2}x-4}~~ = ~~-2x+1} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( \cfrac{1}{2}x-4 \right)~~ = ~~2(-2x+1)}\implies x-8=-4x+2\implies 5x-8=2[/tex]
[tex]5x=10\implies x=\cfrac{10}{5}\implies \blacktriangleright x=2 \blacktriangleleft \\\\\\ \stackrel{\textit{we know that}}{y==\cfrac{1}{2}x-4}\implies y=\cfrac{1}{2}(2)-4\implies \blacktriangleright y = -3 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (2~~,~~-3)~\hfill[/tex]
