What is the solution set of this system of equations?

y = x2 − 3x − 4 x = y + 8 a. {(-1, 0), (4, 0)} b. {(8, 0), (0, 8)} c. {(0, 0)} d. {(2, -6)} e. there is no real solution.

Option D

Step-by-step explanation:

So here we have two equations:

y = x^2 - 3x - 4

x = y + 8

In this case, you have two variables, x, and y, and two linear equations that are independent (one is quadratic and the other linear, so this is easy to see) this means that the system has a solution and is unique.

Here you can try all the given options and find that the solution is option D, but let's do the math.

First, we isolate Y in the second equation and get:

y = x - 8

and now we replace it in the first equation to get:

x- 8 = x^2 - 3x - 4

So now we have the equation:

x^2 - 4x + 4 = 0

Here we see that the determinant of this equation:

b^2 - 4*a*c = 4^2 - 4*1*4 = 0

So we have only one solution:

x = -b/2a = 4/2 = 2

now we know the value of x, we can replace it in the second equation and get:

2 = y + 8

y = 2 - 8 = -6

So the solution is x= 2 and y = -6,

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8