Find the solution set of this system y=x2+3 3x−y+1=0 c. (2,−1,4) d. (2,,4)

d

Step-by-step explanation:

given the 2 equations

y = x² + 3 → (1)

3x - y + 1 = 0 → (2)

substitute y = x² + 3 into (2)

3x - (x² + 3) + 1 = 0

3x - x² - 3 + 1 = 0

-x² + 3x - 2 = 0 ( multiply through by - 1 )

x² - 3x + 2 = 0 ← in standard form

(x - 1)(x - 2) = 0 ← in factored form

equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 2 = 0 ⇒ x = 2

Substitute each value into (1) for corresponding values of y

x = 1 : y = 1 + 3 = 4 → (1, 4)

x = 2 : y = 4 + 3 = 7 → (2, 7)

solutions are (2, 7), (1, 4) → d

f is the midpoint of