Solve the system of equations:

y = 2x^2 - 3

y = 3x - 1

Given:

y = 2x^2 - 3

y = 3x - 1

Equating both:

2x^2 - 3 = 3x - 1

2x^2 - 3x - 3 + 1 = 0

2x^2 - 3x - 2 = 0

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

Substituting both values of x, the possible values of y are:

y = 5 = - 5/2

Therefore, the possible solutions are as follows:

x = 2, y = 5

x = - 1/2, y = - 5/2