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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.**

Challenge

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

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

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 $x+z<y+z.$

Discussion

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 $x−z<y−z.$

Discussion

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 $zx>zy.$

Discussion

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 $zx >zy .$

Pop Quiz

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

Discussion

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

$LHS/5>RHS/5$

$x>-3$

Example

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 $141.$
Discussion

A solution of an inequality is *any* value of the variable that makes the inequality true. As an example, consider the following inequality.

$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. $2x−3<5$

▼

Solve for $x$

AddIneq

$LHS+3<RHS+3$

$2$