6. Solving Inequalities
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Chapter 6
6.
Solving Inequalities
This lesson delves into the methods for solving inequalities, focusing on the properties of inequalities, real numbers, and boundary points. It explains how these properties can be used to isolate variables and find the solution set. For example, the Addition and Subtraction Properties of Inequalities are discussed as tools for simplifying inequalities. The lesson also provides practical scenarios, such as determining the number of seats in a theater or the height requirement for joining a club, to illustrate the application of these mathematical concepts. It emphasizes that inequalities can have multiple or even infinitely many solutions, which can be represented on a number line.
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Lesson Settings & Tools
| | 15 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Inequalities
Slide of 15
Inequalities can be solved like equations. The key step is to isolate the variable on one side of the inequality. Inequalities may have multiple, or even infinitely many, solutions. In this lesson several examples are introduced showing how to use the properties of inequalities to solve them.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Tickets for a Drama: Making Money
Jordan is a member of the drama club at her school. The club plans to stage the play The Little Frog in Town this weekend. They charge $6 per ticket.
a The club hopes to earn at least $400 from the play. If the club has already sold 42 tickets, write an inequality to represent the number of tickets they still need to sell.
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b What is the minimum total number of tickets the club needs to sell to reach their goal?
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Discussion
Properties of Inequalities
Not all inequalities are expressed in the form x<a, x>a, x≤a, or x≥a. Yet, through inverse operations and the Properties of Inequalities, any inequality can be simplified to one of the mentioned forms. Consider the Addition and Subtraction Properties of Inequalities.
Rule
Addition Property of Inequality
Adding the same number to both sides of an inequality generates an equivalent inequality. This equivalent inequality will have the same solution set and the inequality sign remains the same. Let x, y, and z be real numbers such that x<y. Then, the following conditional statement holds true.
If x<y, then x+z<y+z.
Proof
Addition Property of InequalityThe case when x<y will be proven. The remaining cases can be proven similarly. Before starting the proof, the following biconditional statement needs to be considered.
Using the biconditional statement, the last inequality can be rewritten.
x<y⇔y−x>0
Now, the Identity Property of Addition can be applied to the second part of the statement.
y−x>0
IdPropAdd
Identity Property of Addition
y−x+0>0
Rewrite 0 as z−z
y−x+z−z>0
CommutativePropAdd
Commutative Property of Addition
y+z−x−z>0
-a−b=-(a+b)
y+z−(x+z)>0
AddPar
Add parentheses
(y+z)−(x+z)>0
(y+z)−(x+z)>0⇕x+z<y+z
Finally, because x<y, the property is obtained. If x<y, then x+z<y+z.
Discussion
Subtraction Property of Inequality
Subtracting the same number from both sides of an inequality produces an equivalent inequality. The solution set and inequality sign of this equivalent inequality does not change. Let x, y, and z be real numbers such that x<y. Then, the following conditional statement holds true.
If x<y, then x−z<y−z.
Proof
Subtraction Property of InequalityThe case when x<y will be proven. The other cases can be proven using a similar reasoning. Consider the biconditional statement before beginning the proof.
From the biconditional statement, the last inequality can be rewritten.
x<y⇔y−x>0
This property can be proven using the Additive Inverse of z, which is -z. Now, the Identity Property of Addition can be applied to the second part of the statement. y−x>0
IdPropAdd
Identity Property of Addition
y−x+0>0
Rewrite 0 as (-z)−(-z)
y−x+(-z−(-z))>0
CommutativePropAdd
Commutative Property of Addition
y+(-z−(-z))−x>0
RemovePar
Remove parentheses
y−z−(-z)−x>0
-a−b=-(a+b)
y−z−(-z+x)>0
CommutativePropAdd
Commutative Property of Addition
y−z−(x−z)>0
AddPar
Add parentheses
(y−z)−(x−z)>0
(y−z)−(x−z)>0⇕x−z<y−z
Finally, because x<y, the property has been proven. If x<y, then x−z<y−z.
Discussion
Multiplication Property of Inequality
Multiplying both sides of an inequality by a nonzero real number z produces an equivalent inequality. The following conditions about z need to be considered when applying this property.
| Positive z | If z is positive, the inequality sign remains the same. |
|---|---|
| Negative z | If z is negative, the inequality sign needs to be reversed to produce an equivalent inequality. |
For example, let x, y, and z be real numbers such that x<y and z=0. Then, the equivalent inequalities can be written depending on the sign of z.
- If x<y and z>0, then xz<yz.
- If x<y and z<0, then xz>yz.
Proof
Multiplication Property of InequalityThe case when x<y will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.
- x<y if and only if y−x>0.
- If x and y are positive, then xy>0.
- If z is negative, then -z is positive.
Using these properties, the following conditional statements can be proven.
- If x<y and z>0, then xz<yz.
- If x<y and z<0, then xz>yz.
Each conditional statement will be analyzed separately.
When z Is Greater Than 0
It is given that x<y, then using the first property, it is known that y−x is greater than 0.x<y⇔y−x>0
Furthermore, because z>0, from the second property, it can be stated that the product of z and y−x is also greater than 0.
y−x>0z(y−andz>0⇓x)>0
Now, the second part of this conditional statement can be rewritten using the Distributive Property. z(y−x)>0⇔zy−zx>0
From the first property, it can be said that zy−zx>0 if and only if zx<zy. Additionally, because x<y, the conditional statement has been proven. If x<y and z>0, then zx<zy.
When z Is Less Than 0
Again, because x<y, the following statement is valid.x<y⇔y−x>0
Additionally, since z<0, from the third property it follows that -z is positive. Moreover, the product of -z and y−z will be positive. y−x>0-z(y−and-z>0⇓x)>0
Now, -z can be distributed in the second part of the statement.
-z(y−x)>0
▼
Simplify
Distr
Distribute (-z)
(-z)y−(-z)x>0
MultNegPos
(-a)b=-ab
-zy−(-zx)>0
SubNeg
a−(-b)=a+b
-zy+zx>0
AddIneq
LHS+zy>RHS+zy
zx>zy
If x<y and z<0, then zx>zy.
Discussion
Division Property of Inequality
Dividing both sides of an inequality by a nonzero real number z produces an equivalent inequality. However, the following conditions need to be considered.
| Positive z | If z is positive, the inequality sign remains the same. |
|---|---|
| Negative z | If z is negative, the inequality sign needs to be reversed to produce an equivalent inequality. |
For example, let x, y, and z be real numbers such that x<y and z=0. Then, the equivalent inequalities can be written depending on the sign of z.
- If x<y and z>0, then zx<zy.
- If x<y and z<0, then zx>zy.
Proof
Division Property of InequalityThe case when x<y will be proven. The remaining cases can be proven following a similar reasoning. Before starting the proof, the following properties of real numbers need to be considered.
- x<y if and only if y−x is positive.
- If x and z are positive, then zx is also positive.
- If z is negative, then -z is positive.
Using these properties, the following conditional statements can be proven.
- If x<y and z>0, then zx<zy.
- If x<y and z<0, then zx>zy.
Each case will be analyzed separately.
z>0
It is given that x<y, then using the first property, it is known that y−x is greater than 0.x<y⇔y−x>0
Furthermore, because z>0, from the second property, it can be stated that y−x divided by z is also greater than 0.
y−x>0zy−xandz>0⇓>0
Now, the second part of this conditional statement can be rewritten. zy−x>0⇔zy−zx>0
By using the first property, it can be said that zx is less than zy. Additionally, because x<y, the property has been proven. If x<y and z>0, then zx<zy.
z<0
Again, because x<y, the following statement is valid.x<y⇔y−x>0
Additionally, since z<0, from the third property, it follows that -z is positive. Moreover, the quotient of y−x and -z will be positive. y−x>0-zy−xand-z>0⇓>0
Now, the second part of this statement can be rewritten.
-zy−x>0
▼
Simplify
MoveNegDenomToNum
Put minus sign in numerator
z-(y−x)>0
DistrNegSignSwap
-(b−a)=a−b
zx−y>0
WriteDiffFrac
Write as a difference of fractions
zx−zy>0
AddIneq
LHS+zy>RHS+zy
zx>zy
If x<y and z<0, then zx>zy.
Pop Quiz
Identifying the Appropriate Property to Solve the Inequality
In the applet, determine the property used to isolate the variable on one side of the given inequality as shown.
Discussion
Equivalent Inequalities
Two or more inequalities are equivalent inequalities if they have the same solution set. Similar to equivalent equations, applying Properties of Inequality to an inequality produces an equivalent inequality. Consider the following example.
3x+21>6−2x
Three Properties of Inequalities will be used to solve this inequality.
The solution set of every inequality created in this process is x>-3. The Properties of Inequalities produced three equivalent inequalities until the solution set was found.Example
Number of Seats in the Theater Hall
The play The Little Frog in Town received rave reviews from the audience after its first showing. The club decides to put on this play again with one major difference — they will hold it in the greatest theater hall their city has to offer! Now, the drama club needs to know the number of seats in the hall to print new tickets.
4s−132≤s+174
The number of seats b in the balcony is defined by another inequality.
143−3b>b−17
a Find the simplest form of the inequality for the number of ground floor seats.
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b Find the simplest form of the inequality for the number of seats on the balcony.
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c What is the maximum number of seats in the theater hall according to the given inequalities?
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Hint
a Use the properties of inequalities to simplify the given inequality.
b Use the properties of inequalities to simplify the given inequality.
c Use the simplified inequalities from Part A and Part B.
Solution
a Start by examining the given inequality that represents the number of seats on the ground floor.
4s−132≤s+174
Now, use the properties of inequalities to simplify this inequality. Notice that the variable s appears on both sides of the inequality. First, subtract s from both sides of the inequality by using the Subtraction Property of Inequality.
Next, add 132 to both sides of the inequality by using the Addition Property of Inequality.
Finally, divide both sides of the inequality by 3 to get the variable alone by using the Division Property of Inequality.
The simplest form of the given inequality is obtained.
4s−132≤s+174⇕s≤102
This means that the number of seats on the ground floor is less than or equal to 102. In other words, there are at most 102 seats on the ground floor.
b This time, the following inequality will be simplified.
143−3b>b−17
Once again, use the properties of inequalities to simplify it. Start by adding 3b to both sides of the inequality by using the Addition Property of Inequality.
Now, add 17 to both sides of the inequality.
As the last step, divide both sides of the inequality by 4. Note that since 4 is a positive number, the inequality sign stays the same. In the case that the divisor is a negative number, the inequality sign should be reversed.
The simplest form of the given inequality is found.
143−3b>b−17⇕40>b or b<40
This means that 40 is greater than the number of seats on the balcony. In other words there are at most 39 seats on the balcony.
c Recall the simplified forms of the inequalities in Part A and Part B.
Ground Floors≤102 and Balconyb<40
According to these inequalities, there are at most 102 seats on the ground floor and 39 seats on the balcony. Now, add these number of seats to find the maximum number of seats in the theater hall.
102+39=141
The maximum number of seats in the theater hall is 141.
Discussion
Solution Set of an Inequality
A solution of an inequality is any value of the variable that makes the inequality true. As an example, consider the following inequality.
Lastly, the solution set of the inequality can be represented using set-builder notation. 
2x−3<5
Notice that if 0 is substituted for x in the inequality, the inequality holds true. Therefore, it can be said that 0 is a solution to the given inequality.
2(0)−3<?5⇒-3<5 ✓
However, this is not the only value that makes the inequality true. There are other x-values like 1 and 2 that make it true. The set of all possible values that satisfy an inequality is the solution set of an inequality. The solution set can be determined by applying the Properties of Inequalities to isolate the variable on one side of the inequality. 2x−3<5
x<4
Solution Set{x∣x<4}
It is worth noting that the solution set of a linear inequality in one variable can also be represented using a number line. Example
Theater Auditions
After watching excellent theater performances put on by the drama club, so many students want to join it. Several characteristics are highly sought after for new recruits.
-2h+30≤h−315
More importantly, the new recruit needs to get a great score s in their Literature & Language class. The score is defined with the following inequality.
5(s−14)>245
a Find the solution set of the inequality for the appropriate height of a member.
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b Find the solution set of the inequality that represents the score of the literature and language class.
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Hint
a Isolate the variable h by using the properties of inequalities.
b Isolate the variable s by using the properties of inequalities.
Solution
a The solution set of an inequality can be found by isolating the variable on one side of the inequality. Simplify the given inequality by using the properties of inequalities to isolate the variable h.
-2h+30≤h−315
First, subtract 30 from both sides of the inequality using the Subtraction Property of Inequality.
Next, subtract h from both sides of the inequality.
Now, divide both sides of the obtained inequality by -3 to isolate the variable h. Note that according to the Division Property of Inequality, the inequality sign should be reversed since the divisor -3 is a negative number.
-3h≤-345
DivNegIneq
Divide by -3 and flip inequality sign
-3-3h≥-3-345
CalcQuot
Calculate quotient
h≥115
h≥115⇓{h∣h≥115}
This means that this year the drama club is looking for recruits with a height greater than or equal to 115 centimeters.
b Here, the task is to find the solution set of the following inequality.
5(s−14)>245
Once again, proceed in the same way as Part A. Isolate the variable s by using the properties of inequalities. That can be done by dividing both sides of the inequality by 5 to get rid of the factor in front of the parentheses.
5(s−14)>245
DivIneq
LHS/5>RHS/5
55(s−14)>5245
CancelCommonFac
Cancel out common factors
55(s−14)>5245
CalcQuot
Calculate quotient
s−14>49
s>63⇓{s∣s>63}
This means that if someone wants to join the drama club, they needs to get a score greater than 63.
Pop Quiz
Solving Inequalities
In the applet, there are given various different inequalities. Select the correct solution set for the given inequality.
Discussion
Graphing an Inequality on a Number Line
A number line can be used to represent the solution set of an inequality that has one variable. To graph such an inequality, first, determine its type. If it is a strict inequality, then an open boundary point is drawn. Otherwise, a closed boundary point is drawn. Then, the rest of the solution set is shaded accordingly. Consider the following inequality.
x+2<8
The following four steps act as a guide in graphing the given inequality. 1
Determine the Type of Inequality
expand_more The first step is determining if the inequality is strict or non-strict. In this case, the given inequality is strict because the inequality symbol is <.
Strict x+2 < 8
2
Determine the Solution Set and the Boundary Point
expand_more Next, the solution set and the boundary point of the inequality need to be found. This can be done by solving the inequality using the Properties of Inequalities.
Therefore, the boundary point is 6 and the solution set corresponds to all real numbers less than 6.
3
Draw the Boundary Point on the Number Line
expand_more Here, a circle representing the boundary point is drawn on the number line. If the inequality is strict, the circle is open. If the inequality is non-strict, the circle is closed. For this example, the inequality is strict, and the boundary point is 6. Then, an open circle will be drawn on the number line on the number 6.
4
Shade the Rest of the Solution Set
expand_more Finally, the rest of the solution set will be shaded by drawing an arrow that goes along the solution set and starts on the boundary point. For this situation, the solution set corresponds to all numbers less than 6, which means the arrow will be along the left of the boundary point.
It is worth mentioning that the graph of inequalities whose solution sets are all the real numbers are represented with bidirectional arrows that cover all the number line.
Example
Showing the Duration of the Plays on a Number Line
The drama club makes a survey of several audiences about the plays presented during the school year. Then they discussed these surveys' results at the end-of-year meeting.
This year all plays were at least 100 minutes long. Since the audiences found the duration of the plays a bit too long according to the survey results, the club takes a decision to keep the next year's plays less than 90 minutes in length.
a Write an inequality for the duration d of the next year's plays.
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b Choose the graph that describes the solution set of the inequality written in Part A. This will be a way to present the data.
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Hint
a Use a strict inequality to represent the phrase less than.
b The solution set includes real numbers, not only integers.
Solution
a The club decides to keep the duration of the next year's plays below 90 minutes. Let d represent the duration of the plays. Note that less than represent a strict inequality. This means that the inequality sign will be <. With this in mind, write an inequality to show d is less than 90.
d<90
Notice that it can also be written by using the phrasegreater than. In other words, saying d is less than 90 is the same as saying 90 is greater than the duration of the plays d.
d<90⇕90>d
b Now, the graph of the inequality obtained in Part A can be drawn.
d<90
Since the inequality is strict, 90 is not included in the solution set. This means that 90 will be shown with an open circle in the graph. This already eliminates options II and III. Also, notice that the solution set needs to include real numbers, not just integers, because time is a continuous quantity.
Solution Set{d∣ d is a real number less than 90}
The option IV does not make sense for this solution set because it shows only integers. However, Graph I shows exactly the real numbers less than 90. 
This means that the answer is Graph I.
Closure
Finding the Number of Tickets
At the beginning of the lesson, it was given that the drama club charges $6 per ticket and 42 tickets have been already sold. First, calculate the amount of money earned from these 42 tickets.
The number of tickets needs to be an integer. The minimum integer value of x greater than or equal to 24.67 is 25. Now, add this number to 42 to find the total number of tickets.
42⋅6=252
The club earns $252 from selling 42 tickets. Recall that the aim is to earn at least $400 from this play. Note that at least can be represented using the inequality symbol less than or equal to ≤. This will represent the left hand side of the inequality.
400≤
Now, let x be the number of additional tickets to be sold after 42 tickets. Then 6x will represent the amount that will be earned. This means that the total amount will be 252+6x. Recall that this amount needs to be greater than or equal to 400 or in other words, 400 needs to be less than or equal to that amount. With this in mind, now complete the other side of the inequality.
400≤252+6x⇕252+6x≥400
Now, solve the obtained inequality to find the minimum number of tickets sold. As before, use the properties of inequalities to isolate the variable x. Start by subtracting 252 from both sides of the inequality by using the Subtraction Property.
Finally, divide both sides of the inequality by 6 by using the Division Property.
148≤6x
DivIneq
LHS/6≤RHS/6
6148≤66x
CalcQuot
Calculate quotient
24.666666…≤x
RoundDec
Round to 2 decimal place(s)
24.67≤x
42+25=67
The club needs to sell 67 tickets in total to earn at least $400.
Solving Inequalities
Exercise 2.1
Izabella uses the subway to commute to school. A ticket for a single ride costs a student $1.15. The subway company also offers a monthly pass for $45 which gives access to an unlimited number of rides.
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We want to write an inequality that represents how many times Izabella can ride the subway in a month so that a monthly pass is a better deal. We know that a single subway ride costs $ 1.15 and a monthly pass costs $ 45.
| Subway Rides Costs | |
|---|---|
| Single-ride Ticket | $ 1.15 |
| Monthly Pass | $ 45 |
Let x be the number of times Izabella rides the subway in a month. We can find the price of x subway rides on single-ride tickets by multiplying x by the price of a single pass. 1.15x Izabella can also buy a monthly pass and pay $ 45 for x rides. If a monthly pass is a better deal, then it is cheaper than 1.15x. This means that 1.15x is more than 45. Let's write it as an inequality. 1.15x > 45
It is a given that Izabella rides a subway about 50 times per month. In Part A we found the inequality that represents how many times Izabella can ride the subway so that a monthly pass is a better deal than buying single-ride tickets.
1.15x > 45
In the inequality the variable x is the number of times Izabella rides the subway in a month. We will substitute 50 for x and check whether the inequality produces a true statement.
The inequality produces a true statement for x = 50. This means that buying a monthly pass is a better deal than buying single-ride tickets.
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Izabella's mom gives her $55 to go shopping for a dance. She decides buy a dress. It costs $32.
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Izabella can spend up to $ 55 and she wants to buy a dress that costs $ 32. Let x be the remaining amount of money that she can spend after buying the dress. We will write and solve an inequality to find the possible values of x. The total amount of money that she can spend is the sum of the dress price and x. 32 + x We know that she can spend up to $ 55. This means that the total amount of money that she can spend is less than or equal to 55. Let's write it as a non-strict inequality. 32 + x ≤ 55 Next, we will subtract 32 from each side of the inequality to isolate the variable x by using the Subtraction Property of Inequality.
This means that Izabella can spend at most $23. The remaining amount of money after buying the dress can be shown by the following inequalities. 32 + x ≤ 55 ⇕ x ≤ 23
The dress that costs $ 32 is on sale for 15 %. We will find the new price of the dress and rewrite the inequality from Part A. First, let's find the amount of discount by calculating 15 % of 32.
The new price of the dress is the original price minus the amount of the discount. 32- 4.8 = 27.2 We found that the price of the dress on sale is $ 27.20. In Part A we wrote the inequality when the price of the dress is $ 32. The inequality can be used to find the amount of money that Izabella can spend, x. Let's consider the inequality from Part A. 32 + x ≤ 55 The new price of the shirt is $ 27.20. We will write the same inequality but instead of 32, we will write 27.2. 27.2 + x ≤ 55 Notice that we can rearrange the inequality by isolating the variable to solve it. Let's subtract 27.2 from both sides of the inequality.
This means that Izabella can spend at most $27.80 after buying the dress on sale. 27.2 + x ≤ 55 ⇕ x ≤ 27.8
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Students in a science class are divided into 4 equal groups for a project. The teacher says each group should include at least 5 students. Which of the numbers are the possible number of students in the class?
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Students in a science class are divided into 4 equal groups. Let n be the number of students in the science class. We can find the number of students in each group by dividing the total number of students by the number of groups. n/4 We know that in each group there are at least 5 students. This means that n4 should be greater than or equal to 5. Let's write it as an inequality. n/4 ≥ 5 We will solve the inequality to find the possible values of n. Let's multiply both sides of the inequality by 4 by using the Multiplication Property of Inequality.
This means that the number of students should be greater than equal to 20. Notice that the number of students should be a natural number. Let's compare the given numbers with 20. 16 < 20 * 18 < 20 * 20 = 20 ✓ 24 ≥ 20 ✓ 30 ≥ 20 ✓ It was also said that the groups are equal which means they should include equal number of students. Since there are 4 groups, we should consider the numbers that are divisible by 4. With this in mind, let's consider the numbers greater than equal to 20 once again. 20 ÷ 4 =5 ✓ 24 ÷ 4=6 ✓ 30 ÷ 4=7.5 * Since the number of students should be a natural number, there cannot be 7.5 students in a group. The possible numbers are 20 and 24.
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Solving Inequalities
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