To solve the system of equations:
[tex]\[
\left\{\begin{array}{l}
y = -2x - 7 \\
y = x^2 + 6x + 8
\end{array}\right.
\][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
### Step 1: Set the equations equal to each other
Since both equations are equal to [tex]\(y\)[/tex], we can set them equal to each other:
[tex]\[
-2x - 7 = x^2 + 6x + 8
\][/tex]
### Step 2: Arrange into a standard quadratic equation form
Combine all the terms on one side:
[tex]\[
0 = x^2 + 6x + 8 + 2x + 7
\][/tex]
Simplify:
[tex]\[
0 = x^2 + 8x + 15
\][/tex]
### Step 3: Solve the quadratic equation
The quadratic equation to solve is:
[tex]\[
x^2 + 8x + 15 = 0
\][/tex]
We can solve this by factoring. Looking for two numbers that multiply to 15 and add up to 8, we find:
[tex]\[
(x + 3)(x + 5) = 0
\][/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]\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\][/tex]
### Step 4: Find the corresponding [tex]\(y\)[/tex] values
Substitute [tex]\(x = -3\)[/tex] and [tex]\(x = -5\)[/tex] back into either original equation to find the corresponding [tex]\(y\)[/tex] values. We use [tex]\(y = -2x - 7\)[/tex] for simplicity:
For [tex]\(x = -3\)[/tex]:
[tex]\[
y = -2(-3) - 7 = 6 - 7 = -1
\][/tex]
For [tex]\(x = -5\)[/tex]:
[tex]\[
y = -2(-5) - 7 = 10 - 7 = 3
\][/tex]
### Conclusion
The solutions to the system of equations are:
[tex]\[
(x, y) = (-3, -1) \quad \text{and} \quad (x, y) = (-5, 3)
\][/tex]
Therefore, the points of intersection are:
[tex]\[
\boxed{(-5, 3)} \quad \text{and} \quad \boxed{(-3, -1)}
\][/tex]
