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Many situations involve finding the optimal amount of time, money, or a certain quantity of material to fulfill the requirements of some task that could have more than one solution. Some of these situations can be expressed as inequalities. This lesson will explore how to model specific situations as linear inequalities involving one variable and represent their solution set in a number line.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Inequality
Strict vs Non-strict Inequality
Algebraic Expression

Challenge

How Can Real-Life Situations Be Modeled to Help in Making Choices?

Ignacio, has just been offered a paid internship as a junior rocket scientist for a space exploration company. The company has offered Ignacio two options for how to be paid.

Salary options. The first option is $3500 plus 12% of sales. The secon option is $5000 plus 5% of sales

Ignacio is unsure which is the best option. Help him make the best choice by answering the following questions. His salary depends on it!

a Write an inequality representing the earnings from his sales that guarantees option

1

is better than option

2

for Ignacio.

b If Ignacio is sure that he will make at least

$ 35000

worth of sales per month, then which option should he choose?

c Graph the solution set of the inequality found in Part A on a number line.

Explore

Inequalities Come In Different Forms

Inequalities have many forms. Some have the variable on one side and others have them on both sides of the inequality. They can also contain constant terms that are being subtracted from or added to the variable. Examples of these situations can be seen in the following applet.

Think about the following questions.

What happens to an inequality if one number is added to both sides?
What if one number is subtracted from both sides?
How does performing these operations affect the inequality sign?
What happens to the solution set of the inequality?

Discussion

Addition and Subtraction Properties of Inequalities

Similar to equations, inequalities have some properties that allow their manipulation without changing their solution set. When these properties are applied, an equivalent inequality is obtained. The Addition and Subtraction Properties are two of them.

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 $x + z < y + z .$

Rule

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 $x - z < y - z .$

Explore

Inequalities Involving a Variable With a Real Number Coefficient

The Addition and Subtraction Properties of Inequalities can help to isolate the variable on one side of the inequality by creating equivalent inequalities. Although some inequalities can be solved by using these two properties, there are inequalities where the other properties of inequalities need to be used to determine the solution set.

A set of inequalities with the variable term on one side of the inequality

As can be seen, when the variable in an inequality is multiplied by a real number it cannot be isolated on one side of the inequality by using the aforementioned properties. Now, think about the following questions.

What happens to an inequality if both sides are multiplied by the same positive number?
What if both sides are multiplied by a negative number instead?
What if both sides of the inequality are divided by the same positive number?
What if both sides are divided by a negative number?

Discussion

Multiplication and Division Properties of Inequalities

The Addition and Subtraction Properties of Inequalities do just a portion of the work. That is because they do not create the ability to isolate variable terms that contain coefficients. Not to worry, the Multiplication and Division Properties can help in these cases. Together, these properties help to solve inequalities by creating equivalent inequalities.

Rule

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 $z x > z y .$

Rule

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 Some Properties of Inequalities

Knowing which property to use when solving an inequality is of importance because this can minimize mistakes. In the applet, select the property used to produce each equivalent inequality.

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

Discussion

Determining and Representing the Solution of Inequalities

Applying the Properties of Inequalities to one inequality will produce equivalent inequalities. These equivalent inequalities can have a simpler form, making their solutions more straightforward to identify. Since the equivalent inequalities have the same solutions, the solution set of the original inequality can be determined.

Concept

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 + 3

AddTerms

Add terms

2 x < 8

DivIneq

$LHS / 2 < RHS / 2$

x < \frac{8}{2}

CalcQuot

Calculate quotient

x < 4

Lastly, the solution set of the inequality can be represented using set-builder notation.

\underline{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.

Method

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.

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

< .

Strict x + 2 < 8

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.

x + 2 < 8

SubIneq

$LHS - 2 < RHS - 2$

x < 6

Therefore, the boundary point is

6

and the solution set corresponds to all real numbers less than

6 .

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 $6 .$

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.

Example

Using Inequalities to Help a Company Explore Outer Space

An object must travel at a speed of at least $11.2$ kilometers per second to escape Earth's gravitational field. At Gravitasi Z, engineers built a rocket and were tasked with the mission of exploring celestial objects faraway from Earth. However, there is a big problem: The rocket was made to travel at a speed of only $8$ kilometers per second!

Gravitasi Z's engineers plan to improve the rocket in order to accomplish its mission. Solve the following predicaments to help them succeed.

a They need an inequality expressing the speed

s

that should be added to the rocket to surpass the Earth's gravitational field. Write this inequality.

b The engineers wonder by how much, at the least, should the rocket's speed be increased to overcome the gravitational force? Help them find this speed.

c A few junior engineers are staring at some graphs, not quite sure of their meaning.

Number lines representing the possible solution sets of the inequality

They need to know which of the graphs describes the solution set of the inequality. Let them know which is the correct graph.

Hint

a Begin by writing an expression for the final speed of the rocket.

b Solve the inequality found in Part A. The boundary point represents the minimum speed that is missing.

c Draw the boundary point on a number line. Then, identify which side of the boundary point represents the solution set.

Solution

a It is known that the rocket can reach a speed of

8

kilometers per second. Let

s

represent the additional speed that the rocket will gain after improvements. Then, the sum of

s

and

8

will be the speed of the rocket after the improvements.

\underline{Final Speed of the Rocket} s + 8

The company needs the final speed to be at least

11.2

kilometers per second. The phrase at least means greater than or equal to. Therefore, the inequality is non-strict and the symbol must be

\geq .

\underline{Inequality} s + 8 \geq 11.2

b It is asked to find the minimum speed that should be added to the rocket, the minimum value of

s .

To do so, the inequality written in Part A should be solved. Using the Subtraction Property of Inequality,

s

can be isolated.

s + 8 \geq 11.2

SubIneq

$LHS - 8 \geq RHS - 8$

s \geq 11.2 - 8

SubTerm

Subtract term

s \geq 3.2

The boundary point is

3.2,

representing the minimum speed that needs to be added to the rocket. This inequality also means that improvements that cause the rocket to speed up greater than or equal to

3.2

kilometers per second will enable the rocket to start its mission.

c The graph of the inequality written in Part A should be determined.

s + 8 \geq 11.2

To graph this inequality on a number line, the first step is to determine its type. It is a non-strict inequality because it involves the symbol

\geq .

Its solution set and boundary point were found in Part B.

$s + 8 \geq 11.2$
Solution Set	Boundary Point
$s \geq 3.2$	$3.2$

Since the inequality is non-strict, a closed circle will be drawn on the number line on its boundary point $3.2 .$

Now, the rest of the solution set needs to be shaded. Because the speed added needs to be greater than or equal to $3.2,$ the region on the right of the boundary point will be shaded.

This corresponds to Graph II.

Example

Using Inequalities to Understand Profits

Ignacio is an employee at Gravitasi Z. He feels unsure about the success of their space exploration, so he decides to diversify his money making opportunities by investing in cryptocurrency. He has found a crypto exchange company that does all of the actual investing work for him.

The exchange company charges a subscription of $$ 299$ plus a commission of $$ 5$ for every cryptocurrency bought. Also, for an additional fee of $$ 35,$ each cryptocurrency can be sold again later. Ignacio wants to determine the number of cryptocurrencies that will make this investment profitable. Fractions of cryptocurrencies can be bought.

a Write an inequality that expresses the number of cryptocurrencies Ignacios needs to buy to make a profit.

b What is the minimum number of cryptocurrencies that Ignacio needs to buy to at least make some profit? Give the minimum number in integer form.

c The investment company presents the following graphs to Ignacio.

Which of the graphs describes the solution set of the inequality?

Hint

a Begin by writing an expression for the total investment. Then, find an expression for the total sales expected after selling all the cryptocurrencies.

b Solve the inequality found in Part A using the Properties of Inequalities.

c Draw the boundary point on a number line. Then, identify which side of the boundary point represents the solution set.

Solution

a Let

c

be the number of cryptocurrencies. Because the company charges a subscription of

$ 299

plus

$ 5

per cryptocurrency bought, an expression for the total investment can be written by adding the product of

5

and

c

to the subscription cost.

\underline{Total Investment} 299 + 5 c

Additionally, it is given that each cryptocurrency can be later sold for

$ 35

each. Therefore, multiplying

c

and

$ 35

will give the total amount recovered after selling all the cryptocurrencies.

\underline{Total Sales} 35 c

Finally, to make a profit, the total sales need to be greater than the total investment. That means the inequality is strict and its symbol is

> .

The inequality representing the number of cryptocurrencies needed to buy to make a profit can be written using this information.

35 c > 299 + 5 c

b To help Ignacio find the minimum number required to expect make some profit, the inequality needs to be solved. To do so, the Properties of Inequalities can be used. In this case, the Subtraction Property of Inequality should be used first.

35 c > 299 + 5 c

SubIneq

$LHS - 5 c > RHS - 5 c$

35 c - 5 c > 299

SubTerm

Subtract term

30 c > 299

DivIneq

$LHS / 30 > RHS / 30$

c > \frac{2 9 9}{3 0}

UseCalc

Use a calculator

c > 9.9 \overline{6}

RoundDec

Round to $2$ decimal place(s)

c > 9.97

The boundary point is about

9.97 .

This means that as long as Ignacio bought more than

9.97

cryptocurrencies, he can expect to make a profit. Therefore, the minimum he needs to buy is

10 .

c Consider the inequality found in Part A.

35 c > 299 + 5 c

To graph the solution set of this inequality, the first step is determining its type. Because the inequality symbol is

>,

it is a strict inequality. The solution set and the boundary point of this inequality were found in the previous part.

$35 c > 299 + 5 c$
Solution Set	Boundary Point
$c > 9.97$	$9.97$

Since the inequality is strict, an open circle will be drawn on the number line on its boundary point

9.97

Now, the rest of the solution set needs to be shaded. Because the number of cryptocurrencies bought needs to be greater than $9.97,$ the region on the right side of the boundary point will be shaded.

This corresponds Graph I.

Example

Inequalities With Only Integer Solutions

Gravitasi Z wants to sale coffee mugs to the families of their astronauts — at a low price, of course. A company that produces mugs wants to charge Gravitasi Z a fixed amount of $$ 185$ for production plus a tax of $$ 5$ per mug. Gravitasi Z plans to sell each mug for $$ 18 .$

The space exploration company, busy with making rockets, decides to hire a financial advisor to make these calculations. They hope to make at least a small profit. Find the answers to the following situations to become Gravitasi Z's trusted financial advisor.

a Write an inequality representing the number of mugs that Gravitasi Z should order that will guarantee they make a profit. Assume that they will sell all of the mugs that are ordered to be produced.

b Gravitasi Z gave a small budget to this project, so they want to invest in the minimum number of mugs that guarantee a profit. What is this minimum number?

c Choose the graph that describes the solution set of the inequality written in Part A. This will be a way to present the data.

Hint

a Begin by writing an expression for the total expenses and the total sales.

b Can a fraction of a mug be sold?

c The graph of an inequality with integer solutions is represented by plotting points on the values representing its solution set.

Solution

a Let

m

be the number of mugs that they will order from the production company. The cost of production is

$ 185

plus a tax of

$ 5

per mug. Then, the total expenses can be expressed as the sum of

185

and the product of

5

and

m .

\underline{Total Expenses} 185 + 5 m

Since each coffee mug can be sold for

$ 18,

assuming that all mugs will be sold, the total sales can be expressed by the product of

18

and the number of mugs

m .

\underline{Total Sales} 18 m

In order for Gravitasi Z to profit from this project, the total expenses should be less than the total sales. Therefore, the relationship between these two quantities can be shown by the inequality symbol

< .

Total Expenses 185 + 5 m < Total Sales 18 m

b Gravitasi Z wants the minimum number of coffee mugs that guarantee that they will make a profit. To do so, the inequality written in the previous part will be solved using the Properties of Inequalities.

185 + 5 m < 18 m

▼

Solve for

m

SubIneq

$LHS - 5 m < RHS - 5 m$

185 < 18 m - 5 m

SubTerm

Subtract term

185 < 13 m

DivIneq

$LHS / 13 < RHS / 13$

\frac{1 8 5}{1 3} < m

UseCalc

Use a calculator

14.230769 \dots < m

RoundInt

Round to nearest integer

14 < m

RearrangeIneq

Rearrange inequality

m > 14

From this inequality, it can be seen that

14

is the boundary point. Recall that

m

represents the number of mugs. Therefore,

m

must be a positive integer. Combining these findings, the solution set of the inequality can be written as follows.

\underline{Solution Set} {x ∣ x is an integer greater than 14}

The minimum number of mugs that Gravitasi Z should order from the production company is

15 .

c To find the answer, the solution set of the inequality will be graphed. The solution set and the boundary point of the inequality were found in the previous part.

$185 + 5 m < 18 m$
Solution Set	Boundary Point
${x ∣ x is an integer greater than 14}$	$14$

Since the inequality is strict, an open circle will be drawn on the number line on its boundary point $14 .$

Now the solution set needs to be shaded on the number line. Since integers represent the number of mugs, only the points representing integers should be plotted on the number line instead of shading the region greater than $14 .$ The points will be closed as they are the part of the solution set.

This corresponds to Graph IV. Congratulations! Gravitasi Z was impressed by this data, and it looks like someone has a new job as financial advisor.

Example

Solving Inequalities to Win a Game

Ignacio is feeling stress from doing so much rocket science, so he goes with his friend Tearik to an escape room adventure! After some tough challenges, they have reached the final level where only one can win! They stand before two giant doors. Each can be unlocked only by solving a riddle.

two doors with that each have a riddle posted

The riddles are as follows.

Ignacio's Riddle
Two less than the product of four and a number is less than one-half of the sum of eight times the number and six.

Tearrik's Riddle
Three less than the product of five and a number is greater than or equal to twice this number plus three times the sum of this number and two.

It seems like each riddle is an inequality! All they have to do now is give the correct number. Answer the following questions to decipher how Ignacio and Tearrik can unlock their respective doors to get out of the escape room.

a Write and graph the inequality that represents the riddle that Ignacio needs to solve.

b Write and graph the inequality that represents the riddle that Tearrik needs to solve.

Answer

a Inequality:

4 x - 2 < \frac{8 x + 6}{2}

Graph:

b Inequality :

5 x - 3 \geq 2 x + 3 (x + 2)

Graph:

Hint

a Break the riddle into three parts. Then, write an algebraic expression for each part. Finally, joint the expressions to create the inequality. If the variable is canceled out and a true statement is reached, all the real numbers are in the solution set of the inequality.

b If the variable is canceled out and a false statement is reached, the inequality has no solution.

Solution

a Ignacio's riddle can be seen as three parts. Each part can be expressed algebraically.

The expression is less than can be written as

< .

Therefore, the riddle implies a strict inequality. If

x

represents the unknown number, then the first part of the riddle can be written as

4 x - 2 .

The last part can be expressed as

\frac{8 x + 6}{2} .

As a result, Ignacio's riddle is the following inequality.

4 x - 2 < \frac{8 x + 6}{2}

This inequality can be solved using the Properties of Inequalities to find a number Ignacio can use to unlock the door. To remove the fraction, the Multiplication Property of Inequality will be applied first. Recall that if a negative number is multiplied on both sides of the inequality, the inequality symbol needs to be reversed.

4 x - 2 < \frac{8 x + 6}{2}

MultIneq

$LHS \cdot 2 < RHS \cdot 2$

2 (4 x - 2) < 8 x + 6

Distr

Distribute $2$

8 x - 4 < 8 x + 6

SubIneq

$LHS - 6 < RHS - 6$

8 x - 4 - 6 < 8 x

SubTerm

Subtract term

8 x - 10 < 8 x

SubIneq

$LHS - 8 x < RHS - 8 x$

- 10 < 0

Note that the variable has been canceled out. However, because

- 10 < 0

is always a true statement, any number will make the inequality true. Therefore, the solution set of the inequality is all the real numbers. This can be represented on a number line by shading all the number line.

Ignacio can unlock the door by giving any number.

b Following a similar reasoning, the inequality for Tearrik's riddle can be found. The statement can also be divided into three parts. Then each part can be expressed into its corresponding algebraic expression.

The expression greater than or equal to can be written as

\geq .

Therefore, the riddle implies a non-stric inequality. Now, let

x

represent the unknown number. Then the first part of the riddle can be written as

5 x - 3 .

The last part can be expressed as

2 x + 3 (x + 2) .

The following inequality corresponds to Tearrik's riddle.

5 x - 3 \geq 2 x + 3 (x + 2)

In a similar way, this inequality can be solved using the Properties of Inequalities to find a number that Tearrik can used to unlock the door.

5 x - 3 \geq 2 x + 3 (x + 2)

Distr

Distribute $3$

5 x - 3 \geq 2 x + 3 x + 6

AddTerms

Add terms

5 x - 3 \geq 5 x + 6

AddIneq

$LHS + 3 \geq RHS + 3$

5 x \geq 5 x + 6 + 3

AddTerms

Add terms

5 x \geq 5 x + 9

SubIneq

$LHS - 5 x \geq RHS - 5 x$

0 \geq 9 \times

Solving the inequality, a false statement has been reached. This means there is not a number that makes the inequality true. Therefore, the inequality has no solution.This can be represented with a number line that has no shading.

This was a trap all along! Tearrik has no way to escape. Ignacio will be the winner of the escape room.

Closure

Applying the Properties of Inequalities to Make the Best Choice

In this lesson, inequalities with one variable were solved using the Properties of Inequalities. With the help of these properties, the best salary option for Ignacio's can be found. Recall the options offered to Ignacio.

A base pay of $$ 3500$ per month plus $12 %$ of sales.
A base pay of $$ 5000$ per month plus $5 %$ of sales.

Now, the answer for the following questions can be found.

a Write an inequality representing the earnings from sales that guarantees option

1

is better than option

2

for Ignacio.

b If Ignacio is sure that he will make at least

$ 35000

on sales per month, then which is the best choice?

c Graph the solution set of the inequality on a number line.

Answer

a Inequality:

3000 + 0.12 s > 5000 + 0.05 s

b Because Ignacio's sales will be more than the boundary point — which is

$ 28571

— he should go for option

1 .

c Graph:

Hint

a Begin by writing an algebraic expression for each salary option. Then, combine them with an inequality symbol.

b Use the Properties of Inequalities to find the boundary point of the inequality.

c Draw the boundary point on a number line. Then, identify which side of the boundary point represents the solution set.

Solution

a First, each option will be written as an algebraic expression. Let

s

be the earning from sales. Then, Ignacio will get

$ 3500

and

12 %

s .

\underline{First Salary Option} 3500 + 0.12 s

In a similar way, Ignacio will get the sum of

5000

and

5 %

s .

\underline{Second Salary Option} 5000 + 0.05 s

To write an inequality, the two options can be joined. In this scenario, since it is desired that option

1

is the better one, the expression for the first salary option should be greater than the other expression. Therefore, the inequality is strict and the symbol is

> .

3000 + 0.12 s > 5000 + 0.05 s

b Ignacio knows that he can make at least

$ 35000

on sales. If this value is in the solution seteichi of the inequality, Ignacio can choose option

1 .

If not, he should go for option

2 .

The solution set of the inequality can be found by using the Properties of Inequalities. This process will begin by applying the Subtraction Property of Inequality.

3000 + 0.12 s > 5000 + 0.05 s

SubIneq

$LHS - 3000 > RHS - 3000$

0.12 s > 5000 - 3000 + 0.05 s

SubTerm

Subtract term

0.12 s > 2000 + 0.05 s

SubIneq

$LHS - 0.05 s > RHS - 0.05 s$

0.12 s - 0.05 s > 2000

SubTerm

Subtract term

0.07 s > 2000

DivIneq

$LHS / 0.07 > RHS / 0.07$

s > \frac{2 0 0 0}{0 . 0 7}

UseCalc

Use a calculator

s > 28571.428571 \dots

RoundInt

Round to nearest integer

s > 28571

This means that option

1

should be preferred as long as at least

$ 28571

worth of sales are made. Therefore, Ignacio should choose option

1

because

$ 35000

is in the solution set.

c Consider the inequality found in Part A.

3000 + 0.12 s > 5000 + 0.05 s

To graph this inequality, its type needs to be determined. Because it has the inequality symbol

>,

it is a strict inequality. In the previous part, its solution set and the boundary point was found.

$3000 + 0.12 s > 5000 + 0.05 s$
Solution Set	Boundary Point
$s > 28571$	$28571$

Since the inequality is strict, an open circle will be drawn on the number line on its boundary point $28571 .$

Finally, the rest of the solution set needs to be shaded. Because the earnings need to be greater than $28571,$ the region on the right side of the boundary point will be shaded.

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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

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

When $z$ Is Greater Than $0$

When $z$ Is Less Than $0$

$z > 0$

$z < 0$