Solve the system of equation
y=2x^2-3
y=3x-1

Since they both equal y
2x^2-3=y=3x-1
2x^2-3=3x-1
subtract 3x from both sides and add 1
2x^2-3x-2=0
factor
(x-2)(2x+1)=0
set each to zero
x-2=0
x=2

2x+1=0
2x=-1
x=-1/2

subsitute
y=3x-1

y=3(2)-1
y=3(-1/2)-1

y=6-1
y=-3/2-2/2

y=5
y=-5/2

so solutions are
x=2 and y=5 or
x=-1/2 and y=-5/2

(x,y)
(2,5) or (-1/2, -5/2)

The third and fourth box.