We have two equations:
y = x ^ 2 + 4x + 3 --- Equation 1
y = 2x + 6 --- Equation 2
We can set the right-hand sides of the two equations equal to each other since they are both equal to y:
x ^ 2 + 4x + 3 = 2x + 6
Subtracting 2x + 6 from both sides, we get:
x ^ 2 + 2x - 3 = 0
We can factor the quadratic equation:
(x + 3)(x - 1) = 0
This gives us two possible solutions:
x + 3 = 0 or x - 1 = 0
Solving for x, we get:
x = -3 or x = 1
Substituting these values back into either Equation 1 or Equation 2, we can solve for y:
When x = -3, y = (-3) ^ 2 + 4(-3) + 3 = 0
When x = 1, y = 2(1) + 6 = 8
Therefore, the solution to the system of equations is (x, y) = (-3, 0) and (1, 8).