11. Solving and Graphing One Variable Inequalities
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Chapter 1
11.
Solving and Graphing One Variable Inequalities
In this lesson, various aspects of solving and graphing one-variable inequalities are covered. It delves into the properties of inequalities and how they can be applied to find solution sets and boundary points. The lesson also discusses real-life scenarios, such as helping a company make salary choices, to illustrate the practical applications of understanding inequalities. It provides a step-by-step guide for solving inequalities, emphasizing the importance of properties like the Addition and Subtraction Properties of Inequalities. The lesson aims to equip the reader with the skills needed to solve inequalities and understand their implications in real-world situations.
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| | 13 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving and Graphing One Variable Inequalities
Slide of 13
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
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.
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.
Proof
Addition Property of InequalityThe 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
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.
Proof
Subtraction Property of InequalityThe 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
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.
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=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.
Proof
Multiplication Property of InequalityThe 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.
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⇔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>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.
When z Is Less Than 0
Again, because x<y, the following statement is valid.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
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=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.
Proof
Division Property of InequalityThe 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.
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⇔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>0zy−xandz>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 less than zy. Additionally, because x<y, the property has been proven. If x<y and z>0, then zx<zy.
z<0
Again, because x<y, the following statement is valid.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−xand-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
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.
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.
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
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. 1
Determine the Type of Inequality
expand_more 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
2
Determine the Solution Set and the Boundary Point
expand_more 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
expand_more 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
expand_more 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.
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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}
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.
Final Speed of the Rockets+8
The company needs the final speed to be at least 11.2 kilometers per second. The phrase at leastmeans greater than or equal to. Therefore, the inequality is non-strict and the symbol must be ≥.
Inequalitys+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
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.
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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.
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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}
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.
Total Investment299+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 Sales35c
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
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.
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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}
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.
Total Expenses185+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 Sales18m
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 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∣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+5m<18m | |
|---|---|
| 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.
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: 4x−2<28x+6
Graph:
b Inequality : 5x−3≥2x+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.
4x−2<28x+6
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. 4x−2<28x+6
MultIneq
LHS⋅2<RHS⋅2
2(4x−2)<8x+6
Distr
Distribute 2
8x−4<8x+6
SubIneq
LHS−6<RHS−6
8x−4−6<8x
SubTerm
Subtract term
8x−10<8x
SubIneq
LHS−8x<RHS−8x
-10<0
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.
5x−3≥2x+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. 5x−3≥2x+3(x+2)
Distr
Distribute 3
5x−3≥2x+3x+6
AddTerms
Add terms
5x−3≥5x+6
AddIneq
LHS+3≥RHS+3
5x≥5x+6+3
AddTerms
Add terms
5x≥5x+9
SubIneq
LHS−5x≥RHS−5x
0≥9×
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+0.12s>5000+0.05s
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% of s.
First Salary Option3500+0.12s
In a similar way, Ignacio will get the sum of 5000 and 5% of s.
Second Salary Option5000+0.05s
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+0.12s>5000+0.05s
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+0.12s>5000+0.05s
SubIneq
LHS−3000>RHS−3000
0.12s>5000−3000+0.05s
SubTerm
Subtract term
0.12s>2000+0.05s
SubIneq
LHS−0.05s>RHS−0.05s
0.12s−0.05s>2000
SubTerm
Subtract term
0.07s>2000
DivIneq
LHS/0.07>RHS/0.07
s>0.072000
UseCalc
Use a calculator
s>28571.428571…
RoundInt
Round to nearest integer
s>28571
c Consider the inequality found in Part A.
3000+0.12s>5000+0.05s
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+0.12s>5000+0.05s | |
|---|---|
| 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.
Solving and Graphing One Variable Inequalities
Exercises
Consider the following inequality.
8x−14>ax+36
Determine the value of a that makes the inequality have the solution where x>10.
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To determine the value of a, we will use the Properties of Inequalities to isolate the variable term on one side of the inequality.
The Division Property of Inequality can be applied now to isolate the variable. However, we must keep in mind that 8−a must be greater than 0 to keep the inequality symbol. Doing so will make it possible to find the value of a such that the solution set is x>10. x(8-1)>50 ⇔ x>50/8-a Note that we want the right-hand side of the inequality to be equal to 10. This means we can equate 508-a to 10 and solve the resulting equation for a.
Therefore, the value of a that makes the solution set of the inequality be x>10 is 3. 8x-14> 3x+36
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A motorcycle instructor recommends that beginner riders keep a speed of less than 70 miles per hour.
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We are told that the speed limit of a beginner's motorcycle is 70 miles per hour. This means that they have a speed r less than 70 miles per hour. r<70 To determine the distance traveled, we can use a variation of the distance formula, d=rt. d=rt ⇔ r=d/t Using this information, we can write and inequality for d. r<70 Substitute d/t<70 Since the time of practice riding is in minutes, we need to rewrite either the speed limit or the practice time so they both have the same time units. In this case, we will convert 55 minutes to hours using a conversion factor. 1h/60min Using this conversion factor, we can find how many hours correspond to 55 minutes.
Substituting this value for t into the previous inequality will result in the inequality that represents the possible distances d in miles that a beginner rider can travel in 55 minutes of practice time. d/t<70 Substitute d/1112<70
Consider the inequality found in the previous part. d/1112<60 Solving this inequality will give us the distances the rider can travel in 55 minutes of practice time. We can do this by using the Properties of Inequalities. In this case, we need to use the Multiplication Property of Inequality. Let's do it!
This means that a beginner rider can only travel less than 64.17 miles in 55 minutes of practice riding.
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If the circumference C of a circle is known, its radius can be calculated by the formula r=2πC. Consider the following circle.
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We are asked to write an inequality that represents the possible circumference C of a circle whose radius r is greater than 9. Consider the given formula. r=C/2π We will substitute C2π for r in the inequality given in the diagram. r>9 Substitue C/2π>9 This inequality represents the possible circumference C of a circle whose radius is greater than 9.
Consider the inequality found previously. C/2π>9 The possible values for a circumference of a circle with radius greater than 9 can be determined by solving the inequality. To do so, we will use the Multiplication Property of Inequality.
This means that a circle whose radius is greater than 9 will have a circumference that is greater than 18π.
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Hiring an electrician from Company 1 costs $250 per hour plus an initial fee of $100. At Company 2, it costs $150 per hour plus an initial amount of $500. After how many hours x, is it cheaper to hire an electrician from Company 2? Answer with an inequality.
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We will begin by writing an expression for the cost of hiring an electrician for each company.
Company 1
Let x be the number of hours the electrician works. Because Company 1 charges an initial fee of $100 plus 250 per hour worked, adding the product of 100 and x to the initial fee gives the total cost of Company 1. Cost of Company1 100+250x
Company 2
Similarly, Company 2 charges an initial amount of $500 plus $150 per hour of work. This means the total cost of Company 2 can be determined by adding the product of 150 and x to the initial fee of $500. Cost of Company2 500+150x
Writing the Inequality
We want to know after how many hours is it cheaper to hire an electrician from Company 2. Therefore, the expression of the total cost of Company 2 should be less than the expression of the total cost of Company 1. This can be represented with a strict inequality and its is symbol is <. 500+150x<100+250x Let's solve the inequality to find the number of hours required to have a cheaper cost by hiring Company 2.
Company 2 will be cheaper when the work takes more than 4 hours.
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Solving and Graphing One Variable Inequalities
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