To find the solution of the system, we need to set the values of y in both equations equal to each other and solve for x.
So, we have:
x^2 + 3x - 4 = 2x + 2
Simplifying the equation:
x^2 + 3x - 4 - 2x - 2 = 0
x^2 + x - 6 = 0
Now, we can factor the equation:
(x + 3)(x - 2) = 0
Setting each factor equal to zero:
x + 3 = 0 or x - 2 = 0
Solving for x:
x = -3 or x = 2
Now that we have the values of x, we can substitute them back into one of the original equations to find the corresponding y-values.
Using the first equation:
y = x^2 + 3x - 4
For x = -3:
y = (-3)^2 + 3(-3) - 4
y = 9 - 9 - 4
y = -4
For x = 2:
y = (2)^2 + 3(2) - 4
y = 4 + 6 - 4
y = 6
So, the solution of the system is (-3, -4) and (2, 6).
Therefore, the correct answer is b. (negative 3, 4) and (2, 6)
Find the solutions of the system.
y equals x squared plus 3 x minus 4
y equals 2 x plus 2
(1 point)
a. (negative 3, 6) and (2, negative 4)
b. (negative 3, 4) and (2, 6)
c. (negative 3, negative 4) and (negative 2, negative 2)
d. no solution
1 answer