Solve the system of equations.
y = 2x² – 3
y = 3x – 1
(1 point)
Responses
no solution
no solution
(–one-half, 5), (2, –start fraction 5 over 2 end fraction)
(– Image with alt text: one-half , 5), (2, – Image with alt text: start fraction 5 over 2 end fraction )
(–one-half, –start fraction 5 over 2 end fraction), (2, 5)
(– Image with alt text: one-half , – Image with alt text: start fraction 5 over 2 end fraction ), (2, 5)
(one-half, start fraction 5 over 2 end fraction), (2, 5)
To solve the system of equations y = 2x² - 3 and y = 3x - 1, we can set them equal to each other since they both equal y:
2x² - 3 = 3x - 1
Rearranging the equation, we get:
2x² - 3x - 2 = 0
Now we can factor or solve this quadratic equation using the quadratic formula. Factoring, we get:
(2x + 1)(x - 2) = 0
Setting each factor to zero gives us two possible solutions:
2x + 1 = 0
x = -1/2
and
x - 2 = 0
x = 2
Now, substitute these values back into one of the original equations to find the corresponding y values:
For x = -1/2:
y = 2(-1/2)² - 3
y = 2(1/4) - 3
y = 1/2 - 3
y = -5/2
For x = 2:
y = 2(2)² - 3
y = 2(4) - 3
y = 8 - 3
y = 5
Therefore, the solution to the system of equations is (-1/2, -5/2) and (2, 5).