Solve the system: y = x2 + 7, y = 4x + 3Question 39Select one:a.(0, 7)b.(0, 7) and (0, 3)c.(2, 11)d.No solutione.None of these

To solve the system of equations, we need to set the two equations equal to each other because they both equal y.

So, we have:

x^2 + 7 = 4x + 3

Rearranging the terms, we get:

x^2 - 4x + 7 - 3 = 0

which simplifies to:

x^2 - 4x + 4 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it by using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a).

In this case, a = 1, b = -4, and c = 4.

Substituting these values into the quadratic formula, we get:

x = [4 ± sqrt((-4)^2 - 414)] / (2*1) x = [4 ± sqrt(16 - 16)] / 2 x = [4 ± sqrt(0)] / 2 x = [4 ± 0] / 2 x = 2

Substituting x = 2 into the first equation y = x^2 + 7, we get:

y = 2^2 + 7 y = 4 + 7 y = 11

So, the solution to the system of equations is (2, 11), which corresponds to option c.