Answer :
First, we find the equation of the function B:
The equation is of the form y = mx + b, where m is:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]We have the points P1(0,2) and P2(4,0), so
[tex]m=\frac{0-2}{4-0}=-\frac{2}{4}=-\frac{1}{2}[/tex]And b is:
[tex]\begin{gathered} y=mx+b \\ \text{we use the point (0,2)} \\ 2=-\frac{1}{2}(0)+b \\ b=2 \end{gathered}[/tex]Therefore, the function B is:
[tex]y=-\frac{1}{2}x+2[/tex]Next, we solve the system of equations:
[tex]\begin{gathered} y=-3x+2 \\ y=-\frac{1}{2}x+2 \end{gathered}[/tex]Substitute y:
[tex]-3x+2=-\frac{1}{2}x+2[/tex]Solve for x:
[tex]\begin{gathered} -3x+\frac{1}{2}x+2=-\frac{1}{2}x+\frac{1}{2}x+2 \\ -\frac{5}{2}x+2=2 \\ -\frac{5}{2}x+2-2=2-2 \\ -\frac{5}{2}x=0 \\ x=0 \end{gathered}[/tex]Then for y:
[tex]y=-3(0)+2=0+2=2[/tex]Answer:
x = 0
y = 2