5.4: Linear Inequalities in One Variable with Applications
- Page ID
- 129555
Learning Objectives
After completing this section, you should be able to:
- Graph inequalities in one variable.
- Solve linear inequalities in one variable.
- Construct a linear inequality to solve applications.
In this section, we will study linear inequalities in one variable. Inequalities can be used when the possible values (answers) in a certain situation are numerous, not just a few, or when the exact value (answer) is not known but it is known to be within a range of possible values. There are many real-world scenarios that can be represented by linear inequalities. For example, consider the survey of the mayoral election in Figure 5.5 Surveys and polls are usually conducted with only a small group of people. The margin of error indicates a range of how the actual group of voters would vote given the results of the survey. This range can be expressed using inequalities.
Another example involves college tuition. Say a local community college charges $113 per credit hour. You budget $1,500 for tuition this fall semester. What are the number of credit hours that you could take this fall? Since this answer could be many different values, it can be expressed as an inequality.
Graphing Inequalities on the Number Line
In Algebraic Expressions, we introduced equality and the symbol. In this section, we look at inequality and the symbols , , , and . The table below summarizes the symbols and their meaning.
Symbol | Meaning |
---|---|
less than | |
greater than | |
less than or equal to | |
greater than or equal to |
Suppose you had the inequality statement . What possible number or numbers would make the inequality true? If you are thinking, " could be 4," that's correct, but could also be 5, 6, 37, 1 million, or even 3.001. The number of solutions is infinite; any number greater than 3 is a solution to the inequality .
Rather than trying to list all possible solutions, we show all the solutions to the inequality on the number line. All the numbers to the right of 3 on the number line are shaded, to show that all numbers greater than 3 are solutions. At the number 3 itself, an open parenthesis is drawn, since the number 3 is not part of the solutions of .
We can also represent inequalities using interval notation. There is no upper end to the solution to this inequality. In interval notation, we express
We used the left parenthesis symbol to show that the endpoint of the inequality is not included. Parentheses are used when the endpoints are not included as a possible answer to the inequality. The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals.
The inequality
Figure 5.8 summarizes the general representations in both number line form and interval notation of solutions for
Example 5.21
Graphing an Inequality
Graph the inequality
- Answer
Shade to the right of
Figure 5.9)− 3
Write in interval notation starting at
Your Turn 5.21
Example 5.22
Graphing a Compound Inequality
Graph the inequality
- Answer
Step 1: Graph
Figure 5.10).x > − 3
Step 2: Graph
Step 3: Graph both on the same number line and think of where the solutions are to BOTH inequalities Figure 5.12. This will be where BOTH are shaded.
Step 4: Write the solution in interval notation:
Your Turn 5.22
Who Knew?
Where Did the Inequality Symbols Come From?
The first use of the
Solving Linear Inequalities
A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms
When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities. We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example, we know that 2 is less than 4, i.e.,
The same would happen if we subtracted 6 from both sides of the inequality; the statement would stay true:
Notice that the inequality signs stayed the same. This leads us to the Addition and Subtraction Properties of Inequality.
FORMULA
For any numbers
For any numbers
We can add or subtract the same quantity from both sides of an inequality and still keep the inequality the same. But what happens to an inequality when we divide or multiply both sides by a number? Let's first multiply and divide both sides by a positive number, starting with an inequality we know is true,
The inequality signs stayed the same. Does the inequality stay the same when we divide or multiply by a negative number? Let's use our inequality
Notice that when we filled in the inequality signs, the inequality signs reversed their direction in order to make it true! To summarize, when we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses. This gives us the Multiplication and Division Property of Inequality.
FORMULA
For any numbers
multiply or divide by a positive: | if |
if |
|
multiply or divide by a negative: | if |
if |
To summarize, when we divide or multiply an inequality by:
- a positive number, the inequality sign stays the same.
- a negative number, the inequality sign reverses.
Checkpoint
Be careful to only reverse the inequality sign when you are multiplying and dividing by a negative. You do NOT reverse the inequality sign when you add or subtract a negative. For example,
Example 5.23
Solving a Linear Inequality Using One Operation
Solve
- Answer
9 y < 54 9 y 9 < 54 9 y < 6 9 y < 54 9 y 9 < 54 9 y < 6
Your Turn 5.23
Example 5.24
Solving a Linear Inequality Using Multiple Operations
Solve the inequality
- Answer
6 y ≤ 11 y + 17 6 y − 11 y ≤ 11 y + 17 − 11 y − 5 y ≤ 17 − 5 y − 5 ≥ − 17 5 y ≥ − 17 5 6 y ≤ 11 y + 17 6 y − 11 y ≤ 11 y + 17 − 11 y − 5 y ≤ 17 − 5 y − 5 ≥ − 17 5 y ≥ − 17 5
Your Turn 5.24
Solving Applications with Linear Inequalities
Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.
Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.
Example 5.25
Constructing a Linear Inequality to Solve an Application with Tablet Computers
A teacher won a mini grant of $4,000 to buy tablet computers for their classroom. The tablets they would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets the teacher can buy?
- Answer
Let
t = the number of tablets . t = the number of tablets . times $254.12 has to be less than $4,000, sot t .254.12 t ≤ 4 , 000 254.12 t ≤ 4 , 000 Solve for
:t t 254.12 t 254.12 ≤ 4,000 254.12 t ≤ 15.74 254.12 t 254.12 ≤ 4,000 254.12 t ≤ 15.74 The teacher can buy 15 tablets and stay under $4,000.
Your Turn 5.25
Example 5.26
Constructing a Linear Inequality to Solve a Tuition Application
The local community college charges $113 per credit hour. Your budget is $1,500 for tuition this fall semester. What number of credit hours could you take this fall?
- Answer
Let
the number of credit hours you could take.c = c = times $113 has to be less than $1,500, soc c .113 c ≤ 1 , 500 113 c ≤ 1 , 500 Solve for
:c c 113 c 113 ≤ 1500 113 c ≤ 13.27 113 c 113 ≤ 1500 113 c ≤ 13.27 You can take up to 13 credits and stay under $1,500.
Your Turn 5.26
Example 5.27
Constructing a Linear Inequality to Solve an Application with Travel Costs
Brenda’s best friend is having a destination wedding and the event will last 3 days and 3 nights. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 for airfare, $375 for food and entertainment, and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?
- Answer
Let
number of babysitting hours.b = b = times $15 plus $500 has to be more thanb b , so$ 350 + $ 375 + $ 60 / night $ 350 + $ 375 + $ 60 / night .15 b + 500 ≥ 350 + 375 + 60 ( 3 ) 15 b + 500 ≥ 350 + 375 + 60 ( 3 ) Solve for
:b b 15 b + 500 − 500 ≥ 905 − 500 15 b ≥ 405 15 b 15 ≥ 405 15 b ≥ 27 15 b + 500 − 500 ≥ 905 − 500 15 b ≥ 405 15 b 15 ≥ 405 15 b ≥ 27 Brenda must babysit at least 27 hours.
Your Turn 5.27
Tech Check
The Desmos activities called "Inequalities on a Number Line" and "Compound Inequalities on a Number Line" are ways for students to develop and deepen their understanding of inequalities. Teachers will need a Desmos account to assign the activity for student use. Once they have assigned the activity to their students, teachers need to share the code for the activity with their students. Students will input the code to work on the activity.
Check Your Understanding
- /**/[ - 1,\infty )/**/
- /**/{ ( - 1,1)}/**/
- /**/{ (\infty ,1)}/**/
- /**/{ ( - \infty ,1)}/**/
- /**/{ ( - \infty , - 1)}/**/
- /**/( - 5,\infty )/**/
- /**/{ [ - 5,\infty )}/**/
- /**/{ [ - 5,\infty )}/**/
- /**/{ [ - 5, - 3)}/**/
- /**/{ [ - 5, - 3]}/**/
- /**/(1,\infty )/**/
- /**/{ [1,\infty )}/**/
- /**/{ \left[ {\frac{3}{2},\infty } \right)}/**/
- /**/{ \left( {\frac{3}{2},\infty } \right)}/**/
- /**/{ \left( {\infty ,\frac{3}{2}} \right)}/**/
- /**/{( - 4,3)}/**/
- /**/{ (3, - 4)}/**/
- /**/{[ - \infty ,\infty )}/**/
- /**/{ [ - 4,3]}/**/
- /**/{ [3, - 4]}/**/
-
/**/4x \geq 0/**/
-
/**/4x \leq 0/**/
-
/**/6x < 24/**/
-
24" class=" math-rendered">/**/6x > 24/**/
-
/**/6x \geq 24/**/
-
/**/- 6x < 18/**/
-
18" class=" math-rendered">/**/- 6x > 18/**/
-
/**/- 6x \leq 18/**/
-
/**/- 6x \geq 18/**/
-
/**/- 6x \leq - 18/**/
-
- 8" class=" math-rendered">/**/- 4x > - 8/**/
-
- 11" class=" math-rendered">/**/4x + 3 > - 11/**/
-
- 11" class=" math-rendered">/**/4x - 3 > - 11/**/
-
/**/- 4x \leq - 8/**/
-
/**/- 4x + 3 \leq 5/**/
-
/**/9x < 0/**/
-
/**/- 3x \geq 27/**/
-
- 13" class=" math-rendered">/**/- 3x + 14 > - 13/**/
-
- 13" class=" math-rendered">/**/- 3x - 14 > - 13/**/
-
/**/- 3x \geq 27/**/
-
/**/8 < 764x/**/
-
/**/8x < 764/**/
-
764x" class=" math-rendered">/**/8 > 764x/**/
-
764" class=" math-rendered">/**/8x > 764/**/
-
None of these
-
/**/50 < \text{8,120x}/**/
-
/**/50x < \text{8,120}/**/
-
/**/50 < \text{8,120x}/**/
-
\text{8,120}" class=" math-rendered">/**/50x > \text{8,120}/**/
-
None of these