To solve this system of equations, we can substitute y = 2x^2 from the first equation into the second equation for y, and then solve for x:
2x^2 = -3x - 1
Rearranging this equation, we get:
2x^2 + 3x + 1 = 0
Now we can solve for x using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 2, b = 3, and c = 1. Substituting these values, we get:
x = (-3 ± sqrt(3^2 - 4(2)(1))) / 2(2)
x = (-3 ± sqrt(1)) / 4
There are two solutions for x:
x = -1
x = -1/2
Now we can use these values of x to find the corresponding values of y:
When x = -1:
y = 2x^2 = 2(-1)^2 = 2
y = -3x - 1 = -3(-1) - 1 = 2
So one solution to the system of equations is (-1, 2).
When x = -1/2:
y = 2x^2 = 2(-1/2)^2 = 1/2
y = -3x - 1 = -3(-1/2) - 1 = 3/2 - 1 = 1/2
So another solution to the system of equations is (-1/2, 1/2).
Therefore, the solutions to the system of equations are (-1, 2) and (-1/2, 1/2).