Question
Solve the inequality. Graph the solutions.
StartAbsoluteValue StartFraction x plus 15 Over 5 EndFraction EndAbsoluteValue
x+15
5greater than>33
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes in your choice.
A.
The solution is xequals=enter your response here. (Type an integer or a simplified fraction.)
B.
The solution is enter your response hereless than<xless than<enter your response here. (Type integers or simplified fractions.)
C.
The solution is xless than<enter your response here or xgreater than>enter your response here. (Type integers or simplified fractions.)
D.
The solution is all real numbers.
E.
There is no solution.
StartAbsoluteValue StartFraction x plus 15 Over 5 EndFraction EndAbsoluteValue
x+15
5greater than>33
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes in your choice.
A.
The solution is xequals=enter your response here. (Type an integer or a simplified fraction.)
B.
The solution is enter your response hereless than<xless than<enter your response here. (Type integers or simplified fractions.)
C.
The solution is xless than<enter your response here or xgreater than>enter your response here. (Type integers or simplified fractions.)
D.
The solution is all real numbers.
E.
There is no solution.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the inequality
\[
\left| \frac{x + 15}{5} \right| > 33,
\]
we'll start by manipulating the absolute value inequality into two separate inequalities.
1. We can start by setting the fraction inside the absolute value greater than 33:
\[
\frac{x + 15}{5} > 33
\]
2. Next, we also need to consider the case when the fraction inside the absolute value is less than -33:
\[
\frac{x + 15}{5} < -33
\]
Now we will solve each of these inequalities separately.
**For the first inequality:**
\[
\frac{x + 15}{5} > 33
\]
Multiply both sides by 5 (since 5 is positive, the direction of the inequality remains the same):
\[
x + 15 > 165
\]
Subtract 15 from both sides:
\[
x > 150
\]
**For the second inequality:**
\[
\frac{x + 15}{5} < -33
\]
Again, multiplying both sides by 5:
\[
x + 15 < -165
\]
Subtract 15 from both sides:
\[
x < -180
\]
Now we can summarize the solutions. The solution set for the original inequality is:
\[
x < -180 \quad \text{or} \quad x > 150.
\]
Thus, the answer corresponds to Choice C:
**C. The solution is** \(x < -180\) **or** \(x > 150\).
### Graphing the Solutions
The graph of the solutions will consist of two open intervals on a number line:
- An open interval extending left from -180.
- An open interval extending right from 150.
So the number line would look like this:
```
<---|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|--->
-200 -180 -150 -100 -50 0 50 100 150 200
(------) (-------)
```
This represents \( x < -180 \) or \( x > 150 \).
\[
\left| \frac{x + 15}{5} \right| > 33,
\]
we'll start by manipulating the absolute value inequality into two separate inequalities.
1. We can start by setting the fraction inside the absolute value greater than 33:
\[
\frac{x + 15}{5} > 33
\]
2. Next, we also need to consider the case when the fraction inside the absolute value is less than -33:
\[
\frac{x + 15}{5} < -33
\]
Now we will solve each of these inequalities separately.
**For the first inequality:**
\[
\frac{x + 15}{5} > 33
\]
Multiply both sides by 5 (since 5 is positive, the direction of the inequality remains the same):
\[
x + 15 > 165
\]
Subtract 15 from both sides:
\[
x > 150
\]
**For the second inequality:**
\[
\frac{x + 15}{5} < -33
\]
Again, multiplying both sides by 5:
\[
x + 15 < -165
\]
Subtract 15 from both sides:
\[
x < -180
\]
Now we can summarize the solutions. The solution set for the original inequality is:
\[
x < -180 \quad \text{or} \quad x > 150.
\]
Thus, the answer corresponds to Choice C:
**C. The solution is** \(x < -180\) **or** \(x > 150\).
### Graphing the Solutions
The graph of the solutions will consist of two open intervals on a number line:
- An open interval extending left from -180.
- An open interval extending right from 150.
So the number line would look like this:
```
<---|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|--->
-200 -180 -150 -100 -50 0 50 100 150 200
(------) (-------)
```
This represents \( x < -180 \) or \( x > 150 \).
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