Answer:
(- 1, 6 ) and (2, 3 )
Explanation:
y = x² - 2x + 3 → (1)
y = - x + 5 → (2)
substitute y = x² - 2x + 3 into (2)
x² - 2x + 3 = - x + 5 ( add x to both sides )
x² - x + 3 = 5 ( subtract 5 from both sides )
x² - x - 2 = 0 ← in standard form
(x - 2)(x + 1) = 0 ← in factored form
equate each factor to zero and solve for x
x + 1 = 0 ( subtract 1 from both sides )
x = - 1
x - 2 = 0 ( add 2 to both sides )
x = 2
substitute each of these values into (2) for corresponding values of y
x = - 1 : y = - (- 1) + 5 = 1 + 5 = 6 ⇒ (- 1, 6 )
x = 2 : y = - 2 + 5 = 3 ⇒ (2, 3 )
solutions are (- 1, 6 ) and (2, 3 )