To determine the solution set for the given system of equations, let's solve them step by step. The equations are:
1. [tex]\( y + 4 = x^2 \)[/tex]
2. [tex]\( y - x = 2 \)[/tex]
We start with the second equation:
[tex]\[ y - x = 2 \][/tex]
We can solve this for [tex]\( y \)[/tex]:
[tex]\[ y = x + 2 \][/tex]
Now, we substitute [tex]\( y = x + 2 \)[/tex] into the first equation [tex]\( y + 4 = x^2 \)[/tex]:
[tex]\[ (x + 2) + 4 = x^2 \][/tex]
[tex]\[ x + 6 = x^2 \][/tex]
Rearrange this to form a standard quadratic equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Next, we solve this quadratic equation by factoring:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]
So, we find [tex]\( x = 3 \)[/tex] and [tex]\( x = -2 \)[/tex].
Now, we can find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex]:
1. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = x + 2 \][/tex]
[tex]\[ y = 3 + 2 \][/tex]
[tex]\[ y = 5 \][/tex]
Thus, one solution is [tex]\((3, 5)\)[/tex].
2. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = x + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]
So, the other solution is [tex]\((-2, 0)\)[/tex].
Therefore, the solutions to the system of equations are:
[tex]\[ (-2, 0) \quad \text{and} \quad (3, 5) \][/tex]
So, the correct answer is:
[tex]\[ (-2, 0) \text{ and } (3, 5) \][/tex]
