Question

Need help on this question

Answer 1

(-4, -12) and (4, 12)

Explanation:

Since y = 3x AND y = x^2+3x-16,

Substitute x^2+3x-16 in for y in the 1st equation: x^2+3x-16 = 3x;

Solve like you would any other Quadratic Equation:

1. Get all terms on the left: x^2 - 16 = 0

2. Factor if possible: (x-4)(x+4) = 0

3. Set each factor = 0 and solve:

x = 4 and x = -4

4. Substitute each x-solution into one of the original equations to find their y-coordinates: y = 3(-4) = -12 and y = 3(4) = 12

5. List solution(s) as ordered pairs

Answer 2

(- 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 = ±
`√(16)` = ± 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 )