To solve the system of equations:
y = 2x^2 - 3
y = 3x - 1
We can set the two equations equal to each other:
2x^2 - 3 = 3x - 1
Rearranging the equation:
2x^2 - 3x + 2 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation:
a = 2, b = -3, and c = 2
Substituting these values into the quadratic formula:
x = (-(-3) ± √((-3)^2 - 4(2)(2))) / (2(2))
x = (3 ± √(9 - 16)) / 4
x = (3 ± √(-7)) / 4
Since we have a negative value under the square root, the equation has no real solutions. Therefore, the system of equations has no real solutions.