Solve the system of equations y=x^2+3x-4 and y=2x-4

Answer:
A
Step-by-step explanation:
Given the 2 equations
y = x² + 3x - 4 → (1)
y = 2x - 4 → (2)
Substitute y = x² + 3x - 4 into (2)
x² + 3x - 4 = 2x - 4 ( subtract 2x - 4 from both sides )
x² + x = 0 ← in standard form
x(x + 1) = 0 ← in factored form
Equate each factor to zero and solve for x
x = 0
x + 1 = 0 ⇒ x = - 1
Substitute these values into (2) for corresponding values of y
x = 0 → y = 2(0) - 4 = 0 - 4 = - 4 ⇒ (0, - 4 )
x = - 1 → y = 2(- 1) - 4 = - 2 - 4 = - 6 ⇒ (- 1, - 6 )