The equation of the first line is, y=3x-16.
The second line passes through (1,7) and (4,1).
The equation of the second line can be determined as,
[tex]\begin{gathered} \frac{y-7}{x-1}=\frac{7-1}{1-4} \\ y-7=-2(x-1)_{} \\ y=-2x+9 \end{gathered}[/tex]
Substituting the value of y from equation (1) in (2),
[tex]\begin{gathered} 3x-16=-2x+9 \\ 5x=25 \\ x=5 \end{gathered}[/tex]
Substituting the value of x in equation (1),
[tex]\begin{gathered} y=-2(5)+9 \\ y=-1 \end{gathered}[/tex]
Thus, the requried intersection point is (5,-1).