The solution is the intersection point of the 2 lines.
The given two equations are:
[tex]\begin{gathered} y=-\frac{2}{3}x+3 \\ \text{and} \\ y=\frac{3}{2}x-2 \end{gathered}[/tex]
Equating both equations, we can solve for x:
[tex]\begin{gathered} -\frac{2}{3}x+3=\frac{3}{2}x-2 \\ 3+2=\frac{3}{2}x+\frac{2}{3}x \\ 5=\frac{13}{6}x \\ x=\frac{5}{\frac{13}{6}} \\ x=5\times\frac{6}{13} \\ x=\frac{30}{13} \end{gathered}[/tex]
Substituting this value into the second equation, we can find y:
[tex]\begin{gathered} y=\frac{3}{2}x-2 \\ y=\frac{3}{2}(\frac{30}{13})-2 \\ y=\frac{90}{26}-2 \\ y=\frac{45}{13}-\frac{26}{13} \\ y=\frac{19}{13} \end{gathered}[/tex]
The solution set or intersecting point is:
[tex](x,y)=(\frac{30}{13},\frac{19}{13})[/tex]
Converting to decimal and rounding we can say:
(x, y) = (2.3, 1.5)
