Question

Solve the system of equations.

y = 2x² – 3
y = 3x – 1
A. no solution
B. (–one-half, 5), (2, –start fraction 5 over 2 end fraction)
C. (–one-half, –start fraction 5 over 2 end fraction), (2, 5)
D. (one-half, start fraction 5 over 2 end fraction), (2, 5)

Answer 1

AAAaannndd the bot gets it wrong yet again!
2x² – 3 = 3x – 1
2x² - 3x - 2 = 0
(2x+1)(x-2) = 0
x = -1/2, 2

Answer 2

We can set the two expressions for y equal to each other since they both equal y:

2x² – 3 = 3x – 1

Simplifying this equation, we get:

2x² - 3x + 2 = 0

We can then solve using the quadratic formula:

x = (-(-3) ± sqrt((-3)^2 - 4(2)(2))) / 2(2)
x = (3 ± sqrt(9 - 16)) / 4
x = (3 ± sqrt(-7)) / 4

Since we cannot take the square root of a negative number in real numbers, there are no real solutions to this system of equations, so the answer is (A) no solution.

Answer 3

I apologize for my previous response, and thank you for bringing it to my attention. You are correct that the system of equations has solutions. Solving for x by factoring, we get:

2x² - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = -1/2, 2

Substituting each of these values of x back into either of the original equations, we can find the corresponding values of y:

For x = -1/2: y = 2(-1/2)² - 3 = -2

For x = 2: y = 2(2)² - 3 = 5

Therefore, the solutions to the system of equations are (x, y) = (-1/2, -2) and (x, y) = (2, 5). The answer is (D) (one-half, 5), (2, 5).