Solve the following system of equations and show all work: y = x^2 + 4, y = 2x + 1

To solve the system of equations y = x² + 4 and y = 2x + 1, we can set the two equations equal to each other since they both equal y:

x² + 4 = 2x + 1

Now, we will rearrange the equation to set it to zero:

x² – 2x + 4 – 1 = 0

Simplifying this, we get:

x² – 2x + 3 = 0

Next, we will apply the quadratic formula to find the values of x. The quadratic formula is given by:

x = (-b ± √(b² – 4ac)) / 2a

For our equation, a = 1, b = -2, and c = 3. Plugging in these values, we get:

x = (2 ± √((-2)² – 4 * 1 * 3)) / (2 * 1)

x = (2 ± √(4 – 12)) / 2

x = (2 ± √(-8)) / 2

Since the term under the square root, -8, is negative, there are no real solutions for x. This indicates that the solution to the system of equations does not intersect at real points on the graph. Therefore, the system has no solution, meaning the equations represent two curves that do not meet.

We could also express the result in terms of complex numbers, if desired, but for practical purposes in real-valued scenarios, we conclude that the system is inconsistent.