Solve the system of equations
y=2x^2-3
y=3x-1
2x^2-3 = 3x-1
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = -1/2 or x = 2
To solve the system of equations, we can set the two equations equal to each other since they both equal y.
2x^2 - 3 = 3x - 1
Now, let's rearrange the equation to bring all the terms to one side:
2x^2 - 3 - 3x + 1 = 0
Combine like terms:
2x^2 - 3x - 2 = 0
Next, let's solve this quadratic equation. There are a few methods to solve quadratics, but one common method is factoring.
Step 1: Check if the equation can be factored.
The quadratic equation we have, 2x^2 - 3x - 2, cannot be factored easily.
Step 2: Use the quadratic formula to solve the equation.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -3, and c = -2. Substituting these values into the formula:
x = (-(-3) ± √((-3)^2 - 4(2)(-2))) / (2(2))
x = (3 ± √(9 + 16)) / 4
x = (3 ± √(25)) / 4
x = (3 ± 5) / 4
This gives us two possible solutions for x:
x₁ = (3 + 5) / 4 = 8 / 4 = 2
x₂ = (3 - 5) / 4 = -2 / 4 = -1/2
Now, substitute these values of x back into one of the original equations to find the corresponding values of y.
For x = 2:
y = 2(2)^2 - 3
y = 2 * 4 - 3
y = 8 - 3
y = 5
For x = -1/2:
y = 2(-1/2)^2 - 3
y = 2 * 1/4 - 3
y = 1/2 - 3
y = -5/2
Therefore, the solutions to the system of equations are:
(x, y) = (2, 5) and (x, y) = (-1/2, -5/2)