The first linear equation passes through (2,4) and (1,1).
Using two point form the equation can be determined as,
[tex]\begin{gathered} \frac{y-1}{x-1}=\frac{4-1}{2-1} \\ \frac{y-1}{x-1}=\frac{3}{1} \\ y-1=3(x-1) \\ y-1=3x-3 \\ y=3x-2 \end{gathered}[/tex]The second linear equation passes through (2,-2) and (-1,-5).
Using two point form the equation can be determined as,
[tex]\begin{gathered} \frac{y-(-2)}{x-2}=\frac{-5-(-2)}{-1-2} \\ \frac{y+2}{x-2}=\frac{-3}{-3} \\ \frac{y+2}{x-2}=1 \\ y+2=x-2 \\ y=x-4 \end{gathered}[/tex]The system of equation can be solved graphically as,
Thus, (-1,-5) is the required solution.
