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# Applications with Inequalities

## Story problems, set notation, interval notation

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Write and Solve Multi-Step Inequalities Given Problem Situations
License: CC BY-NC 3.0

The marching band at Bernard Middle School always performs during half-time of the football games. The band usually performs different routines that are an average of 8 minutes for each arrangement. During the game, they always perform for at least 42 minutes but not more than 60 minutes. If the average time for each piece is 8 minutes, what is the least number of pieces that the band can perform? What is the greatest number of pieces that the band will perform?

In this concept, you will learn to write and solve multi-step inequalities given problem situations.

### Writing out Multi-Step Inequalities

Two inequalities considered together form a compound inequality. Think about it this way, suppose you know that n>0\begin{align*}n > 0\end{align*} and you also know that n<5\begin{align*}n < 5\end{align*}. In that case, you need to consider those two inequalities together. That means, you need to consider a compound inequality.

There are two types of compound inequalities you should know about.

• A conjunction is a compound inequality that contains the word and. A conjunction is true only if both inequalities are true.
n>0\begin{align*}n>0 \end{align*}and n<5\begin{align*}n<5\end{align*}
• A disjunction is a compound inequality that contains the word or. A disjunction is true if either of the inequalities is true.

You can write compound inequalities when you have words that describe more than one inequality.

Let’s look at an example.

The sum of a number, n\begin{align*}n\end{align*}, and 4 is at least 12 and at most 20.

1. Translate the sentence above into a compound inequality.
2. Solve the compound inequality.

First, consider part a. Translate the sentence into parts for each inequality.

The sum of a number,n,and 4 is at least 12 and at most 20.       n+41220\begin{align*}&\text{The sum of a number}, n, \text{and 4 is at least }12 \text { and at most 20}. \\ &\qquad \qquad \qquad \qquad \qquad \ \downarrow \qquad \qquad \qquad \ \downarrow \qquad \qquad \ \ \ \ \quad \downarrow \\ &\qquad \qquad \qquad \qquad \qquad \ n+4 \qquad \qquad \ge 12 \qquad \qquad \quad \le 20\end{align*}

Look at the original sentence.

The sum of a number, n\begin{align*}n\end{align*}, and 4 is at least 12 and at most 20. The word “and” shows that this sentence represents a conjunction. So, the compound inequality could be written as:

n+412\begin{align*}n+4 \ge 12\end{align*}and n+420\begin{align*}n+4 \le 20\end{align*}

Putting the two inequalities together you get:

12n+420\begin{align*}12 \le n +4 \le 20\end{align*}

Next, consider part b. To solve the compound inequality, treat each inequality separately. Subtract 4 from each side.

n+412n+44124n8n+420n+44204n16\begin{align*}\begin {array}{rcl} & n+4 \ge 12 &&& n+4 \le 20 \\ & n+4 - 4 \ge 12 - 4 &&& n+4-4\le 20 -4 \\ & n \ge 8 &&& n \le 16 \end{array}\end{align*}

So, the solution could be written as: n8\begin{align*} n ≥ 8\end{align*} and n16\begin{align*}n ≤ 16\end{align*}. It could also be written as: 8n16\begin{align*}8 ≤ n ≤ 16\end{align*}.

Let's look at another example.

Brandon earns $7 per hour at his job. The number of hours he works varies from week to week. However, each week he always earns no less than$70 and no more than 140. Let h\begin{align*}h\end{align*} represent the number of hours he works each week. 1. Write a compound inequality to represent this problem situation. 2. Solve the inequality to determine the range in the number of hours Brandon works each week. 3. According to the problem, does Brandon ever work 25 hours in any given week? Explain. First, consider part a. Since Brandon earns7 per hour, you can represent the number of dollars he earns each week by multiplying 7 by the number of hours he works. In other words, you can represent his total earnings as 7×h\begin{align*}7 \times h\end{align*} or 7h\begin{align*}7h\end{align*}.

Now, break the problem into parts and translate each part into an inequality.

each week he always earns no less than $70 and no more than$140     7h 70140\begin{align*}&\ldots \text{each week he always earns no less than } \70 \text{ and no more than } \140 \ldots \\ & \qquad \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \qquad \downarrow \qquad \qquad \qquad \qquad \ \downarrow \\ & \qquad \qquad \qquad \qquad \qquad \ \ 7h \ \quad \quad \ge 70 \qquad \qquad \qquad \quad \le 140\end{align*}

Look at the original sentence. Because of the word and, this compound inequality is an example of a conjunction.

You could represent this conjunction as: 7h70\begin{align*}7h ≥ 70\end{align*} and 7h140\begin{align*}7h ≤ 140\end{align*}. We could also write this conjunction as:

707h140\begin{align*}70 \le 7h \le 140\end{align*}

Next, consider part b.

To solve the compound inequality, you need to treat each inequality separately. Divide each side by 7.

7h707h7707h107h1407h71407h20\begin{align*}\begin{array}{rcl} &7h \ge 70 &&& 7h \le140 \\ & \frac{7h}{7} \ge \frac{70}{7} &&& \frac{7h}{7} \le \frac{140}{7} \\ & h \ge 10 &&& h \le 20 \end{array}\end{align*}

So, the solution could be written as: h10\begin{align*}h ≥ 10\end{align*} and h20\begin{align*}h ≤ 20\end{align*}. It could also be written as: 10h20\begin{align*}10 ≤ h ≤ 20\end{align*}.

This solution shows that the number of hours Brandon works each week is greater than or equal to 10 and less than or equal to 20. In other words, Brandon works 10 to 20 hours in any given week.

Then, consider part c.

In part b, you determined that the number of hours Brandon works in any given week is always less than or equal to 20. This means he never works more than 20 hours in any given week. So, he would not work 25 hours during a particular week.

### Examples

#### Example 1

Earlier, you were given a problem about the music pieces played by the marching band. You need to figure out the range of the number of pieces performed.

First, let’s write down the given information. The average length of each piece is 6 minutes. The band performs for no less than 42 minutes. The band performs for no more than 60 minutes.

Let p\begin{align*}p\end{align*} represent the number of music pieces the band can play. The inequality is:

426p60\begin{align*}42 \le 6p \le 60\end{align*}

Next, solve each inequality for the range. You will have a low range and a high range for our number of pieces.

426p4266p6p76p606p6606p10\begin{align*}\begin{array}{rcl} &42 \le 6p &&& 6p \le 60 \\ &\frac{42}{6} \le \frac{6p}{6} &&& \frac{6p}{6} \le \frac{60}{6}\\ &p \ge 7 &&& p \le 10 \end{array}\end{align*}

The answer is p7\begin{align*}p \ge 7\end{align*}and p10\begin{align*}p \le 10\end{align*} or 7p10\begin{align*}7 \le p \le 10\end{align*}. This means that the band will perform between 7 and 10 pieces.

#### Example 2

Six less than half of a number, x\begin{align*}x\end{align*}, is either less than -1 or greater than 10.

1. Translate the sentence above into a compound inequality.
2. Solve the compound inequality.

First, consider part a. Translate the sentence into parts for each inequality.

Six less than half of a number,x,is either less than -1 or greater than 10.              12x6   <1  >10\begin{align*}& \text{Six less than half of a number}, x, \text{is either less than -1 or greater than 10}.\\ & \qquad \qquad \qquad \ \ \downarrow \qquad \qquad \quad \ \ \ \ \ \ \ \ \qquad \qquad \downarrow \qquad \qquad \qquad \ \ \downarrow \\ &\qquad \qquad \qquad \ \ \frac{1}{2}x-6 \qquad \qquad \qquad \ \ \ <-1 \qquad \qquad \qquad \ \ >10\end{align*}

Look at the original sentence.

Six less than half of a number, x\begin{align*}x\end{align*}, is either greater than 10 or less than -1. The word “or” shows that this sentence represents a disjunction.

So, the compound inequality could be written as:

12x6<1\begin{align*}\frac{1}{2}x-6<-1\end{align*}or 12x6>10\begin{align*}\frac{1}{2}x-6>10\end{align*}

Next, consider part b. To solve the compound inequality, treat each inequality separately. First, add 6 to each side then multiply by 2.

12x6<112x6+6<1+612x<512x6>1012x6+6>10+612x>16\begin{align*}\begin{array}{rcl} &\frac{1}{2}x-6<-1 &&& \frac{1}{2}x-6>10\\ &\frac{1}{2}x-6+6<-1+6 &&& \frac{1}{2}x-6+6>10+6 \\ &\frac{1}{2}x<5 &&& \frac{1}{2}x>16 \end{array}\end{align*}

Then, multiply each side by 2.

12x<52×12x<5×2x<1012x>102×12x>10×2x>20\begin{align*}\begin{array}{rcl} \frac{1}{2}x<5 &&& \frac{1}{2}x>10 \\ 2 \times \frac{1}{2}x<5 \times 2 &&& 2 \times \frac{1}{2}x>10 \times 2 \\ x<10 &&& x>20 \end{array}\end{align*}

The answer could be written as: x<10\begin{align*} x < 10 \end{align*}or x>32\begin{align*} x > 32\end{align*}

#### Example 3

Solve for x\begin{align*}x\end{align*}: The sum of a number and 3 is at least 10 but not more than 25.

First, translate the sentence into parts for each inequality.

The sum of a number and 3 is at least 10 but not more than 25.   x+310   25\begin{align*}&\text{The sum of a number and 3 is at least 10 but not more than 25}. \\ & \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \downarrow \qquad \qquad \qquad \qquad \quad \ \ \downarrow \\ &\qquad \qquad \qquad \quad x+3 \qquad \qquad \ge 10 \qquad \qquad \quad \qquad \ \ \ \le 25\end{align*}

This compound inequality is an example of a conjunction. You could represent this conjunction as: x+310\begin{align*}x + 3 ≥ 10\end{align*} and x+325\begin{align*}x + 3 ≤ 25\end{align*}. We could also write this conjunction as: 10x+325\begin{align*}10 ≤ x + 3 ≤ 25\end{align*}.

Next, to solve the compound inequality, you need to treat each inequality separately. Subtract 3 from both sides.

x+310x+33103x7x+325x+33253x22\begin{align*}\begin{array}{rcl} &x+3 \ge 10 &&& x+3 \le 25 \\ & x+3-3\ge 10 -3 &&& x+3-3 \le 25-3 \\ &x \ge 7 &&& x \le 22 \end{array}\end{align*}

The answer could be written as: x7\begin{align*} x ≥ 7\end{align*} and x22\begin{align*}x ≤ 22\end{align*}. It could also be written as: 7x25\begin{align*}7 ≤ x ≤ 25\end{align*}.

#### Example 4

Solve for x\begin{align*}x\end{align*}: The sum of a number and six is greater than four and less than 12.

First, translate the sentence into parts for each inequality.

The sum of a number and six is greater than four and less than 12. x+6>4<12\begin{align*}& \text{The sum of a number and six is greater than four and less than 12}.\\ & \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \ \downarrow \\ &\qquad \qquad \qquad \quad x+6 \qquad \qquad \qquad \quad >4 \qquad \qquad \qquad <12\end{align*}

This compound inequality is an example of a conjunction. You could represent this conjunction as: x+44\begin{align*}x + 4 ≥ 4\end{align*} and x+412\begin{align*}x + 4 ≤ 12\end{align*}. We could also write this conjunction as: 4x+612\begin{align*}4 ≤ x + 6 ≤ 12\end{align*}.

Next, to solve the compound inequality, you need to treat each inequality separately. Subtract 3 from both sides.

x+6>4x+66>46x>2x+6<12x+66<126x<6\begin{align*}\begin{array}{rcl} &x+6 > 4 &&& x+6<12 \\ & x+6-6>4-6 &&& x+6-6< 12-6 \\ &x > -2 &&& x<6 \end{array}\end{align*}

The answer could be written as: x>2\begin{align*}x > -2\end{align*} and x<6\begin{align*}x < 6\end{align*}. It could also be written as: 2<x<6\begin{align*}-2 < x < 6\end{align*}.

#### Example 5

Solve for x\begin{align*}x\end{align*}: The product of a number times two is greater than fifteen and less than twenty.

First, translate the sentence into parts for each inequality.

The product of a number times two is greater than fifteen and less than twenty.       2x    >15    <20\begin{align*}&\text{The product of a number times two is greater than fifteen and less than twenty}.\\ & \qquad \qquad \qquad \quad \ \downarrow \qquad \qquad \qquad \qquad \ \ \ \quad \downarrow \qquad \qquad \qquad \qquad \ \qquad \downarrow \\ & \qquad \qquad \qquad \quad \ \ 2x \qquad \qquad \qquad \qquad \ \ \ \ >15 \qquad \qquad \qquad \qquad \ \ \ \ <20\end{align*}
This compound inequality is an example of a conjunction. You could represent this conjunction as:

2x>15\begin{align*}2x >15\end{align*} and 2x<20\begin{align*}2x <20\end{align*}.

We could also write this conjunction as:

15<2x<20\begin{align*}15 < 2 x < 20\end{align*}.

Next, to solve the compound inequality, you need to treat each inequality separately. Divide both sides by 2.

2x>152x2>152x>7.52x<202x2<202x<10\begin{align*}\begin{array}{rcl} & 2x > 15 &&& 2x < 20 \\ &\frac{2x}{2} > \frac{15}{2} &&& \frac{2x}{2} < \frac{20}{2} \\ &x > 7.5 &&& x <10 \end{array}\end{align*}

The answer could be written as: x>7.5\begin{align*}x >7.5\end{align*} and x<10\begin{align*}x <10\end{align*}. It could also be written as: 7.5<x<10\begin{align*}7.5 < x <10\end{align*}.

### Review

Use each example to work with compound inequalities.

Two less than a number, x\begin{align*}x\end{align*}, is at least 16 and at most 25

1. Translate the sentence into an inequality.

2. Solve the inequality.

3. Write a mathematical statement to show the range of possibilities for the answer.

One-third of a number, n\begin{align*}n\end{align*}, is either less than -5 or greater than 3.

4. Translate the sentence into an inequality.

5. Solve the inequality.

6. Write a mathematical statement to show the range of possibilities for the answer.

Seven more than twice a number, n\begin{align*}n\end{align*}, is either less than -5 or at least 9.

7. Translate the sentence into an inequality.

8. Solve the inequality.

9. Write a mathematical statement to show the range of possibilities for the answer.

Solve each problem.

When Harriet goes to the diner for lunch, she buys exactly two items: a wrap sandwich and a $3 milkshake. The total cost of her lunch is always more than$8 and less than 12. Let w\begin{align*}w\end{align*} represent the cost, in dollars, of any of the wrap sandwiches Harriet buys. 10. Write a compound inequality to represent this problem situation. 11. Solve the inequality you wrote in part a. 12. According to the problem, is it possible that Harriet sometimes buys a wrap sandwich that costs10?

13. Why? Explain.

Mr. Jameson pays $3 per gallon for gasoline. Each week, the amount he spends on gas for his car is always at least$30 on gas and at most \$105. Let g\begin{align*}g\end{align*} represent the number of gallons of gasoline he buys in any given week.

14. Write a compound inequality to represent this problem situation.

15. Solve the inequality you wrote in part a.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.18.

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### Vocabulary Language: English

Compound Inequality

A compound inequality is an inequality that combines two other inequalities. For example: $5 < x < 10$, meaning $x > 5$ and $x < 10$.

inequality

An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.

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1. [1]^ License: CC BY-NC 3.0

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