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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.**

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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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*.

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 $x+z<y+z.$

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.
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 $x+z<y+z.$

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 $x−z<y−z.$

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.
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 $x−z<y−z.$

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 $xz>yz.$

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 $xy>0.$
- 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.

$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 $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 $zx<zy.$

$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 $zx>zy.$

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 $zx >zy .$

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 $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.

$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 $y−x>0zy−x andz>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 If $x<y$ and $z>0,$ then $zx <zy .$

$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−x and-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 $zx >zy .$

In the applet, determine the property used to isolate the variable on one side of the given inequality as shown.

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.
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.

$3x+21>6−2x $

Three Properties of Inequalities will be used to solve this inequality.
$3x+21>6−2x$

AddIneq

$LHS+2x>RHS+2x$

$5x+21>6$

SubIneq

$LHS−21>RHS−21$

$5x>-15$

DivIneq

$LHS/5>RHS/5$

$x>-3$

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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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.

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>borb<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 $141.$
Inequalities compare two quantities and often involve one or more variables. 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.
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.

For instance, at the moment, the drama club does not have many tall members. For this recruiting year, they have defined a height to consider for new recruits $h$ with the following inequality.$-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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a Isolate the variable $h$ by using the properties of inequalities.

b Isolate the variable $s$ by using the properties of inequalities.

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

$5 5 (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 $63.$
In the applet, there are given various different inequalities. Select the correct solution set for the given inequality.

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. *expand_more*
*expand_more*
*expand_more*
*expand_more*

$x+2<8 $

The following four steps act as a guide in graphing the given inequality. 1

Determine the Type of Inequality

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 $<.$

$Strictx+2<8 $

2

Determine the Solution Set and the Boundary Point

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

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

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.

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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a Use a strict inequality to represent the phrase *less than*.

b The solution set includes real numbers, not only integers.

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 phrase$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 $Solution Set{d∣ dis a real number less than90} $

The option This means that the answer is **Graph I**.

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.
*at least* can be represented using the inequality symbol *less than or equal to* $≤.$ This will represent the left hand side of the inequality.
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 $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.$