Answer:
(- 4, - 12 ) and (4, 12 )
Explanation:
y = 3x → (1)
y = x² + 3x - 16
substitute y = x² + 3x - 16 into (1)
x² + 3x - 16 = 3x ( subtract 3x from both sides )
x² - 16 = 0 ( add 16 to both sides )
x² = 16 ( take square root of both sides )
x = ±
= ± 4
substitute x = - 4 and x = 4 into (1) for corresponding values of y
x = - 4 : y = 3(- 4) = - 12 ⇒ (- 4, - 12 )
x = 4 : y = 3(4) = 12 ⇒ (4, 12 )
solutions are (- 4, - 12 ) and (4, 12 )