Solve the system of equations. show your work. y = x^2 - 3x + 4 and x + y = 4

This is a parabola intercepted by a straight line. It will have 0, 1 or 2 solutions.

If you convert the straight line to y = 4-x, and fill it in to the first one, you get:

4-x = x^2-3x+4
x^2 - 2x = 0
x(x-2) = 0
x = 0 or x=2

Great! We didn't even need the quadratic formula.

Fill in these x values in either formula (or both to double check!) to find that (0,4) and (2,2) are the solutions. Put differently:

x=0, y=4 and x=2, y=2 are the solutions.

The answer to your question is b’(16,14)