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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.**

Challenge

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.

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 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

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

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 $x+z<y+z.$

Rule

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 $x−z<y−z.$

Explore

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.

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

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

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 $zx>zy.$

Rule

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

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.

Discussion

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

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.
Method

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.

Example

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.

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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.

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

{"type":"choice","form":{"alts":["Graph I","Graph II","Graph III","Graph IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}

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.

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.

$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 leastmeans

$Inequality s+8≥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.

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

$s+8≥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 $≥.$ Its solution set and boundary point were found in Part B. $s+8≥11.2$ | |
---|---|

Solution Set | Boundary Point |

$s≥3.2$ | $3.2$ |

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

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.

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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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">c<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["10"]}}

c The investment company presents the following graphs to Ignacio.

{"type":"choice","form":{"alts":["Graph I","Graph II","Graph III","Graph IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

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.

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.

$Total Investment 299+5c $

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.
$Total Sales 35c $

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.
$35c>299+5c $

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.

$35c>299+5c$

SubIneq

$LHS−5c>RHS−5c$

$35c−5c>299$

SubTerm

Subtract term

$30c>299$

DivIneq

$LHS/30>RHS/30$

$c>30299 $

UseCalc

Use a calculator

$c>9.96$

RoundDec

Round to $2$ decimal place(s)

$c>9.97$

c Consider the inequality found in Part A.

Since the inequality is strict, an open circle will be drawn on the number line on its boundary point $9.97$.

$35c>299+5c $

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. $35c>299+5c$ | |
---|---|

Solution Set | Boundary Point |

$c>9.97$ | $9.97$ |

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

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.

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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?

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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.

{"type":"choice","form":{"alts":["Graph I","Graph II","Graph III","Graph IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}

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.

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 $m.$

$Total Expenses 185+5m $

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 $m.$ $Total Sales 18m $

In order for Gravitasi Z to profit from this project, the total expenses should be $Total Expenses185+5m < Total Sales18m $

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.

$185+5m<18m$

$m>14$

$Solution Set {x∣xis an integer greater than14} $

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.

$185+5m<18m$ | |
---|---|

Solution Set | Boundary Point |

${x∣xis an integer greater than14}$ | $14$ |

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.