Question
Solve the absolute value inequality using an equivalent compound inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.
StartAbsoluteValue x minus 5 EndAbsoluteValue minus 4 less than or equals 2
Question content area bottom
Part 1
A.
[negative 1,2]
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at 2 and the region to the right of a left bracket at negative 1 is shaded.
-4
0
4
8
12
16
B.
(negative 1,11)
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right parenthesis at 11 and the region to the right of a left parenthesis at negative 1 is shaded.
-4
0
4
8
12
16
C.
(minusinfinity,negative 1] union [11,infinity)
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at negative 1 and the region to the right of a left bracket at 11 are shaded.
-4
0
4
8
12
16
D.
[negative 1,11]
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at 11 and the region to the right of a left bracket at negative 1 is shaded.
-4
0
4
8
12
16
StartAbsoluteValue x minus 5 EndAbsoluteValue minus 4 less than or equals 2
Question content area bottom
Part 1
A.
[negative 1,2]
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at 2 and the region to the right of a left bracket at negative 1 is shaded.
-4
0
4
8
12
16
B.
(negative 1,11)
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right parenthesis at 11 and the region to the right of a left parenthesis at negative 1 is shaded.
-4
0
4
8
12
16
C.
(minusinfinity,negative 1] union [11,infinity)
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at negative 1 and the region to the right of a left bracket at 11 are shaded.
-4
0
4
8
12
16
D.
[negative 1,11]
An infinite number line, labeled from negative 7 to 17, has tick marks in increments of 1. The region to the left of a right bracket at 11 and the region to the right of a left bracket at negative 1 is shaded.
-4
0
4
8
12
16
Answers
Answered by
GPT-4o mini
To solve the absolute value inequality \( |x - 5| - 4 \leq 2 \), we start by rearranging it:
1. Add 4 to both sides:
\[
|x - 5| \leq 6
\]
2. Now we can translate the absolute value inequality into a compound inequality:
\[
-6 \leq x - 5 \leq 6
\]
3. Next, we solve this compound inequality. We can break it down into two parts:
- For the left part:
\[
-6 \leq x - 5 \implies -6 + 5 \leq x \implies -1 \leq x \implies x \geq -1
\]
- For the right part:
\[
x - 5 \leq 6 \implies x \leq 6 + 5 \implies x \leq 11
\]
4. Combining both parts, we have:
\[
-1 \leq x \leq 11
\]
This can be expressed in interval notation as:
\[
[-1, 11]
\]
Now, let's graph this solution on a number line:
- You would place a closed bracket at -1 and a closed bracket at 11, and shade the region between them to indicate all numbers in that range.
Based on your answer choices, the correct answer is:
**A.** \([-1, 11]\)
### Graph:
On the number line, it would look like this:
```
<----|----|----|----|----|----|----|----|----|----|----|----|---->
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 11 12 13 14 15
[=================================]
```
This indicates all numbers from -1 to 11, inclusive.
1. Add 4 to both sides:
\[
|x - 5| \leq 6
\]
2. Now we can translate the absolute value inequality into a compound inequality:
\[
-6 \leq x - 5 \leq 6
\]
3. Next, we solve this compound inequality. We can break it down into two parts:
- For the left part:
\[
-6 \leq x - 5 \implies -6 + 5 \leq x \implies -1 \leq x \implies x \geq -1
\]
- For the right part:
\[
x - 5 \leq 6 \implies x \leq 6 + 5 \implies x \leq 11
\]
4. Combining both parts, we have:
\[
-1 \leq x \leq 11
\]
This can be expressed in interval notation as:
\[
[-1, 11]
\]
Now, let's graph this solution on a number line:
- You would place a closed bracket at -1 and a closed bracket at 11, and shade the region between them to indicate all numbers in that range.
Based on your answer choices, the correct answer is:
**A.** \([-1, 11]\)
### Graph:
On the number line, it would look like this:
```
<----|----|----|----|----|----|----|----|----|----|----|----|---->
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 11 12 13 14 15
[=================================]
```
This indicates all numbers from -1 to 11, inclusive.