Let's solve the given system of equations step by step:
[tex]\[
\begin{cases}
y = 2x + 1 \\
y = x^2 + 2x - 8
\end{cases}
\][/tex]
First, recognize that both equations are set equal to [tex]\(y\)[/tex]. This fact allows us to set the right-hand sides of the equations equal to each other:
[tex]\[ 2x + 1 = x^2 + 2x - 8 \][/tex]
Next, we will solve this equation for [tex]\(x\)[/tex]. Subtract [tex]\(2x + 1\)[/tex] from both sides to get:
[tex]\[ 0 = x^2 + 2x - 8 - 2x - 1 \][/tex]
Simplify:
[tex]\[ 0 = x^2 - 9 \][/tex]
This is a quadratic equation, which can be factored:
[tex]\[ 0 = (x - 3)(x + 3) \][/tex]
Setting each factor to zero gives us the solutions for [tex]\(x\)[/tex]:
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
Now we have two values for [tex]\(x\)[/tex]. To find the corresponding [tex]\(y\)[/tex] values, we substitute each [tex]\(x\)[/tex] value back into one of the original equations. Using [tex]\(y = 2x + 1\)[/tex]:
1. For [tex]\(x = 3\)[/tex]:
[tex]\[ y = 2(3) + 1 = 6 + 1 = 7 \][/tex]
So, one solution is [tex]\((3, 7)\)[/tex].
2. For [tex]\(x = -3\)[/tex]:
[tex]\[ y = 2(-3) + 1 = -6 + 1 = -5 \][/tex]
So, the other solution is [tex]\((-3, -5)\)[/tex].
Therefore, the solutions to the system of equations are [tex]\((-3, -5)\)[/tex] and [tex]\((3, 7)\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{A. (-3,-5) \text{ and } (3,7)} \][/tex]
