Answer:
(- 2, - 3 ) and (2, 5 )
Explanation:
given the system of equations
y = x² + 2x - 3 → (1)
y = 2x + 1 → (2)
substitute y = x² + 2x - 3 into (2)
x² + 2x - 3 = 2x + 1 ( subtract 2x + 1 from both sides )
x² - 4 = 0 ← x2 - 4 is a difference of squares
(x + 2)(x - 2) = 0 ← in factored form
equate each factor to zero and solve for x
x + 2 = 0 ⇒ x = - 2
x - 2 = 0 ⇒ x = 2
substitute these value for x into (2) for corresponding values of y
x = - 2 : y = 2(- 2) + 1 = - 4 + 1 = - 3 ⇒ ( - 2, - 3 )
x = 2 : y = 2(2) + 1 = 4 + 1 = 5 ⇒ (2, 5 )
solutions are (- 2, - 3 ) and (2, 5 )