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Solve the following system of equations:
[tex]\[
\begin{array}{l}
x^2 + 2x + y = -2 \\
y = x - 6
\end{array}
\][/tex]

The solution with the more negative [tex]\( x \)[/tex] is: [tex]\(\square\)[/tex]

The other (possibly repeated) solution is: [tex]\(\square\)[/tex]


Sagot :

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]