Asked by hudson
Solve the absolute value inequality −2|x−7|≤−8
. Provide the solution in interval notation.(1 point)
The solution to −2|x−7|≤−8
is ([
]∪
[
]).
Remaining Attempts : 3
. Provide the solution in interval notation.(1 point)
The solution to −2|x−7|≤−8
is ([
]∪
[
]).
Remaining Attempts : 3
Answers
Answered by
GPT-4o mini
To solve the inequality \(-2|x - 7| \leq -8\), we first isolate the absolute value expression:
\[
|x - 7| \geq 4
\]
This is achieved by dividing both sides of the inequality by -2, and remember that dividing by a negative number reverses the inequality sign.
Now, we translate the absolute value inequality \(|x - 7| \geq 4\) into two separate inequalities:
1. \(x - 7 \geq 4\)
2. \(x - 7 \leq -4\)
Now, we solve both inequalities.
1. For \(x - 7 \geq 4\):
\[
x \geq 11
\]
2. For \(x - 7 \leq -4\):
\[
x \leq 3
\]
Now, we combine these results into interval notation. The solutions are:
- \(x \leq 3\) corresponds to the interval \((- \infty, 3]\)
- \(x \geq 11\) corresponds to the interval \([11, \infty)\)
Thus, the solution in interval notation is:
\[
(-\infty, 3] \cup [11, \infty)
\]
So the final answer is:
\[
\text{The solution to } -2|x - 7| \leq -8 \text{ is } \left( (-\infty, 3] \cup [11, \infty) \right).
\]
\[
|x - 7| \geq 4
\]
This is achieved by dividing both sides of the inequality by -2, and remember that dividing by a negative number reverses the inequality sign.
Now, we translate the absolute value inequality \(|x - 7| \geq 4\) into two separate inequalities:
1. \(x - 7 \geq 4\)
2. \(x - 7 \leq -4\)
Now, we solve both inequalities.
1. For \(x - 7 \geq 4\):
\[
x \geq 11
\]
2. For \(x - 7 \leq -4\):
\[
x \leq 3
\]
Now, we combine these results into interval notation. The solutions are:
- \(x \leq 3\) corresponds to the interval \((- \infty, 3]\)
- \(x \geq 11\) corresponds to the interval \([11, \infty)\)
Thus, the solution in interval notation is:
\[
(-\infty, 3] \cup [11, \infty)
\]
So the final answer is:
\[
\text{The solution to } -2|x - 7| \leq -8 \text{ is } \left( (-\infty, 3] \cup [11, \infty) \right).
\]