Applying Properties of Inequalities to Find Solution Sets of Different Inequalities

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.

Number Line
Inverse Operations
Inequality
Strict and Non-strict Inequalities
Solution Set of an Inequality
Writing and Graphing Inequalities

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.

b What is the minimum total number of tickets the club needs to sell to reach their goal?

Discussion

Properties of Inequalities

Not all inequalities are expressed in the form $x < a,$ $x > a,$ $x \leq a,$ or $x \geq a .$ 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 $x + z < y + z .$

This property holds for the other types of inequalities.

Proof

Addition Property of Inequality

The 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.

x < y \Leftrightarrow 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

Using the biconditional statement, the last inequality can be rewritten.

(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 $x - z < y - z .$

This property holds for the other types of inequalities.

Proof

Subtraction Property of Inequality

The 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.

x < y \Leftrightarrow 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

From the biconditional statement, the last inequality can be rewritten.

(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 $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

This property holds for the other types of inequalities.

Proof

Multiplication Property of Inequality

The 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 $x y > 0 .$
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 $x z < y z .$
If $x < y$ and $z < 0,$ then $x z > y z .$

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 \Leftrightarrow 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 > 0 z (y - and z > 0 ⇓ x) > 0

Now, the second part of this conditional statement can be rewritten using the Distributive Property.

z (y - x) > 0 \Leftrightarrow z y - z x > 0

From the first property, it can be said that

z y - z x > 0

if and only if

z x < z y .

Additionally, because

x < y,

the conditional statement has been proven.

x < y

and

z > 0,

then

z x < z y .

When $z$ Is Less Than $0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow 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

$(- a) b = - a b$

- z y - (- z x) > 0

SubNeg

$a - (- b) = a + b$

- z y + z x > 0

AddIneq

$LHS + z y > RHS + z y$

z x > z y

Finally, because

x < y,

the property has been proven.

If $x < y$ and $z < 0,$ then $z x > z y .$

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 $z \neq = 0 .$ Then, the equivalent inequalities can be written depending on the sign of $z .$

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

This property holds for the other types of inequalities.

Proof

Division Property of Inequality

The 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 $\frac{x}{z}$ 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 $\frac{x}{z} < \frac{y}{z} .$
If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

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 \Leftrightarrow 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 > 0 \frac{y - x}{z} and z > 0 ⇓ > 0

Now, the second part of this conditional statement can be rewritten.

\frac{y - x}{z} > 0 \Leftrightarrow \frac{y}{z} - \frac{x}{z} > 0

By using the first property, it can be said that

\frac{x}{z}

is less than

\frac{y}{z} .

Additionally, because

x < y,

the property has been proven.

If $x < y$ and $z > 0,$ then $\frac{x}{z} < \frac{y}{z} .$

$z < 0$

Again, because

x < y,

the following statement is valid.

x < y \Leftrightarrow 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 \frac{y - x}{- z} and - z > 0 ⇓ > 0

Now, the second part of this statement can be rewritten.

\frac{y - x}{- z} > 0

▼

Simplify

MoveNegDenomToNum

Put minus sign in numerator

\frac{- ( y - x )}{z} > 0

DistrNegSignSwap

$- (b - a) = a - b$

\frac{x - y}{z} > 0

WriteDiffFrac

Write as a difference of fractions

\frac{x}{z} - \frac{y}{z} > 0

AddIneq

$LHS + \frac{y}{z} > RHS + \frac{y}{z}$

\frac{x}{z} > \frac{y}{z}

Finally, because

x < y,

the property has been obtained.

If $x < y$ and $z < 0,$ then $\frac{x}{z} > \frac{y}{z} .$

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.

An applet showing different inequalities and its equivalent inequality that results of applying one of the properties of inequalities

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.

3 x + 21 > 6 - 2 x

Three Properties of Inequalities will be used to solve this inequality.

3 x + 21 > 6 - 2 x

AddIneq

$LHS + 2 x > RHS + 2 x$

5 x + 21 > 6

SubIneq

$LHS - 21 > RHS - 21$

5 x > - 15

DivIneq

$LHS / 5 > RHS / 5$

x > - 3

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.

The hall consists of two floors, the ground floor and the balcony. The number of seats

s

on the ground floor is defined by the following inequality.

4 s - 132 \leq s + 174

The number of seats

b

in the balcony is defined by another inequality.

143 - 3 b > b - 17

a Find the simplest form of the inequality for the number of ground floor seats.

b Find the simplest form of the inequality for the number of seats on the balcony.

c What is the maximum number of seats in the theater hall according to the given inequalities?

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.

4 s - 132 \leq 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.

4 s - 132 \leq s + 174

SubIneq

$LHS - s \leq RHS - s$

4 s - 132 - s \leq s + 174 - s

SubTerms

Subtract terms

3 s - 132 \leq 174

Next, add

132

to both sides of the inequality by using the Addition Property of Inequality.

3 s - 132 \leq 174

AddIneq

$LHS + 132 \leq RHS + 132$

3 s - 132 + 132 \leq 174 + 132

AddTerms

Add terms

3 s \leq 306

Finally, divide both sides of the inequality by

3

to get the variable alone by using the Division Property of Inequality.

3 s \leq 306

DivIneq

$LHS / 3 \leq RHS / 3$

\frac{3 s}{3} \leq \frac{3 0 6}{3}

CalcQuot

Calculate quotient

s \leq 102

The simplest form of the given inequality is obtained.

4 s - 132 \leq s + 174 ⇕ s \leq 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 - 3 b > b - 17

Once again, use the properties of inequalities to simplify it. Start by adding

3 b

to both sides of the inequality by using the Addition Property of Inequality.

143 - 3 b > b - 17

AddIneq

$LHS + 3 b > RHS + 3 b$

143 - 3 b + 3 b > b - 17 + 3 b

AddTerms

Add terms

143 > 4 b - 17

Now, add

17

to both sides of the inequality.

143 > 4 b - 17

AddIneq

$LHS + 17 > RHS + 17$

143 + 17 > 4 b - 17 + 17

AddTerms

Add terms

160 > 4 b

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.

160 > 4 b

DivIneq

$LHS / 4 > RHS / 4$

\frac{1 6 0}{4} > \frac{4 b}{4}

CalcQuot

Calculate quotient

40 > b

The simplest form of the given inequality is found.

143 - 3 b > 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 Floor s \leq 102 and Balcony b < 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.

2 x - 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 \Rightarrow - 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.

2 x - 3 < 5

▼

Solve for

x

AddIneq

$LHS + 3 < RHS + 3$

2 x < 5 +

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Catch-Up and Review

Proof

Proof

Proof

When z Is Greater Than 0

When z Is Less Than 0

Proof

z>0

z<0

Hint

Solution

When $z$ Is Greater Than $0$

When $z$ Is Less Than $0$

$z > 0$

$z < 0$