To solve the given system of equations:
[tex]\[
\begin{cases}
x^2 + 2x + y = -2 \\
y = x - 6
\end{cases}
\][/tex]
we can follow these steps:
1. Substitute the second equation into the first equation:
The second equation is [tex]\(y = x - 6\)[/tex]. We substitute [tex]\(y\)[/tex] in the first equation:
[tex]\[
x^2 + 2x + (x - 6) = -2
\][/tex]
2. Simplify the equation:
Combine like terms:
[tex]\[
x^2 + 2x + x - 6 = -2
\][/tex]
[tex]\[
x^2 + 3x - 6 = -2
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
Move [tex]\(-2\)[/tex] to the left side of the equation:
[tex]\[
x^2 + 3x - 6 + 2 = 0
\][/tex]
[tex]\[
x^2 + 3x - 4 = 0
\][/tex]
This is a quadratic equation. To solve it, we factor the quadratic:
[tex]\[
x^2 + 3x - 4 = (x + 4)(x - 1) = 0
\][/tex]
Setting each factor to zero gives the solutions for [tex]\(x\)[/tex]:
[tex]\[
x + 4 = 0 \quad \text{or} \quad x - 1 = 0
\][/tex]
[tex]\[
x = -4 \quad \text{or} \quad x = 1
\][/tex]
4. Find the corresponding [tex]\(y\)[/tex]-values:
Using the second equation [tex]\(y = x - 6\)[/tex], we find the [tex]\(y\)[/tex]-values corresponding to each [tex]\(x\)[/tex]-value:
- For [tex]\(x = -4\)[/tex]:
[tex]\[
y = -4 - 6 = -10
\][/tex]
- For [tex]\(x = 1\)[/tex]:
[tex]\[
y = 1 - 6 = -5
\][/tex]
5. Identify the solutions:
We have the pairs [tex]\((x, y)\)[/tex]:
[tex]\[
(-4, -10) \quad \text{and} \quad (1, -5)
\][/tex]
Among these solutions, the [tex]\(x\)[/tex]-value of [tex]\(-4\)[/tex] is more negative than [tex]\(1\)[/tex].
So, the solution with the more negative [tex]\(x\)[/tex] is:
[tex]\[
\boxed{-4}
\][/tex]
The other (possibly repeated) solution is:
[tex]\[
\boxed{1}
\][/tex]
