To solve the inequality [tex]\(\frac{30}{x + 6} \leq 5\)[/tex], we will follow a step-by-step method:
1. First, isolate the fraction:
[tex]\[
\frac{30}{x + 6} \leq 5
\][/tex]
2. Eliminate the fraction by multiplying both sides by [tex]\(x + 6\)[/tex]:
We need to consider two cases, because the sign of [tex]\(x + 6\)[/tex] affects the inequality:
- When [tex]\(x + 6 > 0\)[/tex] (i.e., [tex]\(x > -6\)[/tex])
- When [tex]\(x + 6 < 0\)[/tex] (i.e., [tex]\(x < -6\)[/tex])
### Case 1: [tex]\(x + 6 > 0\)[/tex] (i.e., [tex]\(x > -6\)[/tex])
In this case, multiplying both sides by [tex]\(x + 6\)[/tex] (a positive number) does not change the direction of the inequality:
[tex]\[
30 \leq 5(x + 6)
\][/tex]
[tex]\[
30 \leq 5x + 30
\][/tex]
Subtract 30 from both sides:
[tex]\[
0 \leq 5x
\][/tex]
[tex]\[
0 \leq x
\][/tex]
or equivalently:
[tex]\[
x \geq 0
\][/tex]
Since we are considering the case [tex]\(x + 6 > 0\)[/tex], we have [tex]\(x \geq 0\)[/tex].
### Case 2: [tex]\(x + 6 < 0\)[/tex] (i.e., [tex]\(x < -6\)[/tex])
In this case, multiplying both sides by [tex]\(x + 6\)[/tex] (a negative number) reverses the direction of the inequality:
[tex]\[
30 \geq 5(x + 6)
\][/tex]
[tex]\[
30 \geq 5x + 30
\][/tex]
Subtract 30 from both sides:
[tex]\[
0 \geq 5x
\][/tex]
[tex]\[
0 \geq x
\][/tex]
or equivalently:
[tex]\[
x \leq 0
\][/tex]
However, in this scenario we have [tex]\(x < -6\)[/tex], which is already included in the region we need to consider.
### Combine the two cases:
- From Case 1, we have [tex]\(x \geq 0\)[/tex].
- From Case 2, we have [tex]\(x < -6\)[/tex].
Since these results represent mutually exclusive conditions, we express this in interval notation:
### Solution set:
[tex]\[
x < -6 \text{ or } x \geq 0
\][/tex]
In interval notation, this is written as:
[tex]\[
(-\infty, -6) \cup [0, \infty)
\][/tex]
This is the final solution set for the inequality [tex]\(\frac{30}{x + 6} \leq 5\)[/tex].
