To solve the inequality [tex]\(\frac{-2}{7-x} \geq \frac{1}{x+1}\)[/tex], we need to follow multiple steps to determine the intervals where the inequality holds true.
### Step-by-Step Solution:
1. Identify the Domain:
The first step is to determine the domain of the inequality. The expressions [tex]\(\frac{-2}{7-x}\)[/tex] and [tex]\(\frac{1}{x+1}\)[/tex] have denominators, which must not be zero.
- For [tex]\(\frac{-2}{7-x}\)[/tex], [tex]\(7-x \neq 0 \Rightarrow x \neq 7\)[/tex].
- For [tex]\(\frac{1}{x+1}\)[/tex], [tex]\(x+1 \neq 0 \Rightarrow x \neq -1\)[/tex].
Thus, the domain excludes [tex]\(x = 7\)[/tex] and [tex]\(x = -1\)[/tex].
2. Combine the Fractions:
Bring the inequality to a common denominator to combine the fractions:
[tex]\[
\frac{-2}{7-x} - \frac{1}{x+1} \geq 0
\][/tex]
The common denominator is [tex]\((7-x)(x+1)\)[/tex]. Rewriting the inequality:
[tex]\[
\frac{-2(x+1) - 1(7-x)}{(7-x)(x+1)} \geq 0
\][/tex]
3. Simplify the Expression:
Simplify the numerator:
[tex]\[
-2(x+1) - (7-x) = -2x - 2 - 7 + x = -x - 9
\][/tex]
So, the inequality becomes:
[tex]\[
\frac{-x - 9}{(7-x)(x+1)} \geq 0
\][/tex]
4. Determine Critical Points:
To determine when [tex]\(\frac{-x - 9}{(7-x)(x+1)}\)[/tex] changes sign, find the critical points by setting the numerator and the denominator to zero:
- Numerator: [tex]\(-x - 9 = 0 \Rightarrow x = -9\)[/tex]
- Denominator: [tex]\(7-x = 0 \Rightarrow x = 7\)[/tex] and [tex]\(x+1 = 0 \Rightarrow x = -1\)[/tex]
The critical points are [tex]\(x = -9\)[/tex], [tex]\(x = -1\)[/tex], and [tex]\(x = 7\)[/tex].
5. Create a Sign Chart:
Use the critical points to divide the real number line into intervals and test the sign of the expression in each interval:
- Intervals: [tex]\((-\infty, -9)\)[/tex], [tex]\((-9, -1)\)[/tex], [tex]\((-1, 7)\)[/tex], and [tex]\((7, \infty)\)[/tex].
6. Test Each Interval:
Evaluate the sign of [tex]\(\frac{-x - 9}{(7-x)(x+1)}\)[/tex] in each interval:
- For [tex]\(x \in (-\infty, -9)\)[/tex]: Choose [tex]\(x = -10\)[/tex]
[tex]\[
\frac{-(-10) - 9}{(7+10)(-10+1)} = \frac{1}{(17)(-9)} < 0
\][/tex]
- For [tex]\(x \in (-9, -1)\)[/tex]: Choose [tex]\(x = -5\)[/tex]
[tex]\[
\frac{-(-5) - 9}{(7+5)(-5+1)} = \frac{4}{(12)(-4)} < 0
\][/tex]
- For [tex]\(x \in (-1, 7)\)[/tex]: Choose [tex]\(x = 0\)[/tex]
[tex]\[
\frac{-0 - 9}{(7-0)(0+1)} = \frac{-9}{7} < 0
\][/tex]
- For [tex]\(x \in (7, \infty)\)[/tex]: Choose [tex]\(x = 8\)[/tex]
[tex]\[
\frac{-8 - 9}{(7-8)(8+1)} = \frac{-17}{(-1)(9)} = \frac{17}{9} > 0
\][/tex]
7. Combine the Results:
From the intervals tested, the expression [tex]\(\frac{-x - 9}{(7-x)(x+1)}\)[/tex] is greater than or equal to zero only in the interval [tex]\((7, \infty)\)[/tex].
8. Check the Endpoint Inclusion:
Only values that satisfy non-strict inequalities ([tex]\(\geq\)[/tex]) can be included:
- [tex]\(x = -9\)[/tex] is included because the numerator becomes zero, satisfying [tex]\(\frac{-x - 9}{(7-x)(x+1)} = \frac{0}{(7-x)(x+1)} = 0\)[/tex].
- [tex]\(x = -1\)[/tex] and [tex]\(x = 7\)[/tex] are excluded from the domain.
### Final Answer:
[tex]\[
\boxed{(-9, -1) \cup (7, \infty)}
\][/tex]
