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)