Question

EE

Solve the system of equations.
y= 2x + 4
y = x2 + x - 2
A. (-2, 0) and (3, 10)
B. (2,0) and (-3,-10)
C. (-3,-10) and (-2,0)
D. (-3, 10) and (2,0)

Answer 1

A

Explanation:

Given the 2 equations

y = 2x + 4 → (1)

y = x² + x - 2 → (2)

Substitute y = x² + x - 2 into (1)

x² + x - 2 = 2x + 4 ( subtract 2x + 4 from both sides )

x² - x - 6 = 0 ← in standard form

(x - 3)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 3 = 0 ⇒ x = 3

x + 2 = 0 ⇒ x = - 2

Substitute these values into (1) for corresponding values of y

x = 3 : y = 2(3) + 4 = 6 + 4 = 10 ⇒ (3, 10 )

x = - 2 : y = 2(- 2) + 4 = - 4 + 4 = 0 ⇒ (- 2, 0 )