Solve the system of equations y=4x+1, y=x2+2x-2

Option D

(-1,-3) and (3,13)

Given in the question two equations

y = 4x + 1

y = x² + 2x - 2

Equate both functions

4x + 1 = x² + 2x - 2

rearrange the x term and constant

-x² + 4x - 2x + 2 + 1 = 0

-x² + 2x + 3 = 0

factors

-x * 3x = - 3x²

-x + 3x = 2x

-x² - x + 3x + 3 = 0

-x (x+1) + 3 (x+1) = 0

solve

(x+1) (3-x) = 0

x = - 1

and

x = 3

Plug these values in equation to find y

x = - 1

y = 4 (-1) + 1

y = - 3

x = 3

y = 4 (3) + 1

y = 13

Option d is the answer to your question