Solve the system: x = y2 + 4y – 5, x = 0.Question 17Select one:a.(0, –5) and (1, 1)b.(4, 4) and (–5, –5)c.(0, 1) and (0, –5)d.No solutione.None of these

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

So, we have:

y^2 + 4y - 5 = 0

This is a quadratic equation, which we can solve using the quadratic formula: y = [-b ± sqrt(b^2 - 4ac)] / (2a)

In this equation, a = 1, b = 4, and c = -5.

Substituting these values into the quadratic formula gives:

y = [-4 ± sqrt((4)^2 - 41(-5))] / (2*1) y = [-4 ± sqrt(16 + 20)] / 2 y = [-4 ± sqrt(36)] / 2 y = [-4 ± 6] / 2

This gives us two solutions: y = 1 and y = -5.

Substituting y = 1 into the equation x = y^2 + 4y - 5 gives x = 0.

Substituting y = -5 into the equation x = y^2 + 4y - 5 also gives x = 0.

So, the solutions to the system of equations are (0, 1) and (0, -5), which corresponds to answer choice c.