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.
Show less Show more expand_more| Student Learning Objectives: |
|---|
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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 problems have more than one solution, so rather than finding the answer, the goal is to find the best answer. 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.
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Challenge
How Can Real-Life Situations Be Modeled to Help when Making Choices?
Ignacio has just been offered a paid internship as a junior rocket scientist for a space exploration company. The company has offered him 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.
a
Write an inequality for Ignacio's sales earnings that guarantees Option 1 is better than Option 2 for Ignacio.
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b
If Ignacio is sure that he will make at least $35 000 worth of sales per month, which option should he choose?
{"type":"choice","form":{"alts":["Option 1","Option 2"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
c
Graph the solution set of the inequality found in Part A on a number line.
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Inequalities Come In Different Forms
Inequalities have many forms. Some have variables on one side and others have them on both sides of the inequality. They can also contain constant terms.
| Example Inequalities | ||
|---|---|---|
| 85≤ 12g-35 | 104f-72> -46 | 60b-98≤ 52 |
| 72+87x<63x | 13-6s≥-47s+54 | 81-u<31+3u |
Consider 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?
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Discussion
Addition and Subtraction Properties of Inequalities
Similar to equations, inequalities have some properties that allow their manipulation without changing their solution set. Applying the properties creates an equivalent inequality. 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 InequalityThe case when x< y will be proven. Before starting the proof, consider the following biconditional statement.
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
The last inequality can be rewritten using the biconditional statement. (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 InequalityThe case when x< y will be proven. Consider the biconditional statement before beginning the proof.
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
The last inequality can be rewritten using the biconditional statement. (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.
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Inequalities Involving a Variable With a Real Number Coefficient
The Addition and Subtraction Properties of Inequalities can help to isolate a 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.
- 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?
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Discussion
Multiplication and Division Properties of Inequalities
The Addition and Subtraction Properties of Inequalities do just a portion of the work because they do not create the ability to isolate variable terms that contain coefficients. 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.
This property holds for the other types of inequalities.
Proof
Multiplication Property of InequalityThe case when x
- 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
As it is given that x< y, then by 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>0 &and z>0 &⇓ z(y-&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
If x0, then zx
When z Is Less Than 0
The following statement is valid because x
- 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
Finally, the property has been proven because x
If x
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
- If x
0, then xz< yz. - If x
yz.
This property holds for the other types of inequalities.
Proof
Division Property of InequalityThe case when x
- x
if and only if y-x is positive. - If x and z are positive, then xz is also positive.
- If z is negative, then - z is positive.
The following conditional statements can be proven using these properties.
- If x
0, then xz< yz. - If x
yz.
Each case will be analyzed separately.
z>0
As it is given that x
If x
z<0
The following statement is valid because x
y-x/- z>0
▼
Simplify
MoveNegDenomToNum
Put minus sign in numerator
-(y-x)/z>0
DistrNegSignSwap
-(b-a)=a-b
x-y/z>0
WriteDiffFrac
Write as a difference of fractions
x/z-y/z>0
AddIneq
LHS+y/z>RHS+y/z
x/z>y/z
Finally, the property has been obtained because x
If x
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Pop Quiz
Identifying Some Properties of Inequalities
Knowing which property to use when solving an inequality is important because it can minimize mistakes. In the applet, select the property used to produce each equivalent inequality.
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Discussion
Determining and Representing the Solution of Inequalities
Applying the Properties of Inequalities to one inequality will produce equivalent inequalities. These equivalent inequalities often 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 using the simpler inequality.
Concept
Solution Set of an Inequality
A solution of an inequality is any value of the variable that makes the inequality true. If a value is substituted for the variable and creates a false statement, that value is not in a solution of the inequality. Consider the following inequality when x=-3, 0, and 3. 2x-3< 5 lcl 2( -3)-3? <5 & ⇒ & -9 < 5 ✓ [0.3em] 2( 0)-3? <5 & ⇒ & -3 < 5 ✓ [0.3em] 2( 3)-3? <5 & ⇒ & 6 ≮ 5 * The set of all possible values that satisfy an inequality is the solution set of an inequality. The solution set of the inequality can be represented using set-builder notation. \begin{gathered} \underline\textbf{Solution Set} \\ \{x\,|\,x\lt 4\} \end{gathered} 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.
Consider the following inequality. x+2<8 There are four steps for 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 <. x + 2 < 8_(Strict)
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 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. The inequality is strict for this example, so an open circle will be drawn on the boundary point 6.
4
Shade the Rest of the Solution Set
expand_more Finally, the rest of the solution set is shaded by drawing an arrow that starts on the boundary point and goes along the solution set. For this situation, the solution set is to all numbers less than 6, which means that the arrow will be along the left of the boundary point.
The graph of inequalities whose solution sets are all the real numbers are represented with bidirectional arrows that cover all the number line.
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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 to explore celestial objects far away 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
THe company needs 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
What is the minimum amount that the rocket's speed needs to be increased by to overcome the gravitational force?
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c
A few junior engineers are considering some graphs.
Which graph describes the solution set of the inequality from Part A?
{"type":"choice","form":{"alts":["Graph I","Graph II","Graph III","Graph IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
Hint
a
Start 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 additional speed required.
c
Draw the boundary point on a number line. Then, identify which side of the boundary point represents the solution set.
Solution
a
We know 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 least
means greater than or equal to. Therefore, the inequality is non-strict and the symbol must be ≥.
Inequality s+8≥ 11.2
b
Now we are asked to find the minimum speed that should be added to the rocket, or the minimum value of s. We can do this by solving the inequality we wrote in Part A. Notice that we can isolate s by using the Subtraction Property of Inequality.
The boundary point is 3.2, which represents 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 allow the rocket to escape Earth's gravitational field.
c
Now we want to determine which of the graphs represents the inequality written in Part A.
s + 8 ≥ 11.2 To graph this inequality on a number line, the first step is to determine its type. Our inequality is a non-strict inequality because it involves the symbol ≥. We found its solution set and boundary point 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.
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Example
Using Inequalities to Understand Profits
Ignacio wants t earn extra money by starting a lawn-care business. He must spend $299 on equipment and advertising. He also spends $5 on gas and supplies for every hour he works. Ignacio charges customers $35 per hour. He wants to determine how many hours he must work before his business becomes profitable. He can work fractions of an hour.
a
Write an inequality that expresses the number of hours Ignacio needs to work to make a profit.
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b
What is the minimum whole number of hours Ignacio needs to work to make a profit?
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c
Ignacio charts his profits on a number line.
Which of the graphs describes the solution set of the inequality?
{"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 Ignacio's total expenses. Then, find an expression for the total amount he earns from working.
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 h represent the number of hours Ignacio works. He spends $299 to get started and an additional $5 for supplies for every hour he works, so his total expenses the sum of the equipment and the product of 5 and h.
Total Cost 299+5h Additionally, we know that Ignacio earns $35 per hour. Therefore, multiplying h and $35 will give the total amount he receives for his work.= Total Earned 35h Finally, to make a profit, the total earned need to be greater than the total cost. That means the inequality is strict and its symbol is >. The inequality expresses the number of hours Ignacio needs to work to make a profit can be written using this information. 35h > 299+5h
b
To help Ignacio find the minimum whole number of hours he needs to work to make some profit, we need to solve the inequality using the Properties of Inequalities. In this case, we will use the Subtraction Property of Inequality first.
35h>299+5h
SubIneq
LHS-5h>RHS-5h
35h-5h>299
SubTerm
Subtract term
30h>299
DivIneq
.LHS /30.>.RHS /30.
h>299/30
UseCalc
Use a calculator
h>9.96
RoundDec
Round to 2 decimal place(s)
h>9.97
The boundary point is about 9.97. This means that as long as Ignacio works for longer than 9.97 hours, he can expect to make a profit. Therefore, he needs to work 10 hours at a minimum.
c
Consider the inequality we created in Part A.
35h> 299+5h
| 35h>299+5h | |
|---|---|
| Solution Set | Boundary Point |
| h> 9.97 | 9.97 |
Since the inequality is strict, we will place an open circle on the number line on its boundary point 9.97.
Now the rest of the solution set needs to be shaded. Because the Ignacio needs to work more than 9.97 hours, the region on the right side of the boundary point will be shaded.
This corresponds Graph I.
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Example
Inequalities With Only Integer Solutions
Gravitasi Z wants to sell 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.
a
Write an inequality representing the number of mugs that Gravitasi Z should order that will guarantee they make a profit if all the mugs are sold.
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b
What is this minimum number of mugs that the company needs to sell to make a profit?
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c
Choose the graph that describes the solution set of the inequality from Part A.
{"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 we buy a part of a mug?
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 Gravitasi Z 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 less than the total sales. Therefore, the relationship between these two quantities can be shown by the inequality symbol <. Total Expenses & & Total Sales 185+5m & < & 18m
b
Gravitasi Z wants the minimum number of coffee mugs that guarantees that they will make a profit. We can find it by solving the inequality from Part A using the Properties of Inequalities.
185+5m<18m
SubIneq
LHS-5m
185<18m-5m
SubTerm
Subtract term
185<13m
DivIneq
.LHS /13.<.RHS /13.
185/13
UseCalc
Use a calculator
14.230769...
RoundInt
Round to nearest integer
14
RearrangeIneq
Rearrange inequality
m>14
From this inequality, we can see that 14 is the boundary point. Recall that m represents the number of mugs purchased and sold. We can only buy and sell mugs in whole number increments, so we can write the solution set to reflect this. 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, the fist integer greater than 14.
c
We want to identify the graph that represents the solution set of the inequality. The solution set and the boundary point of the inequality were found in Part B.
| 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 the mugs can only be sold in whole number increments, instead of shading the continuous region greater than 14, we will only plot the points representing integers. These points will be closed as they are the part of the solution set.
This corresponds to Graph IV.
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Example
Solving Inequalities to Win a Game
Ignacio and his friend Tearik go to an escape room adventure. After some tough challenges, they have reached the final level where only one person can win. They stand in front of 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. 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.
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b
Write and graph the inequality that represents the riddle that Tearrik needs to solve.
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Answer
a
Inequality: 4x-2<8x+6/2
Graph:
b
Inequality : 5x-3≥2x+3(x+2)
Graph:
Hint
a
If the variable is canceled out and a true statement is reached, all 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 broken down into three parts that can each be expressed algebraically.
The expression is less than can be written as <, so the riddle implies a strict inequality. If x represents the unknown number, then the first part of the riddle can be written as 4x-2. The last part can be expressed as 8x+62.
We can combine them to create an inequality that represents Ignacio's riddle. 4x-2<8x+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, we will first use the Multiplication Property of Inequality. Recall that if a negative number is multiplied on both sides of the inequality, the inequality symbol needs to be reversed.
4x-2<8x+6/2
MultIneq
LHS* 2 < RHS * 2
2(4x-2)<8x+6
Distr
Distribute 2
8x-4<8x+6
SubIneq
LHS-6
8x-4-6<8x
SubTerm
Subtract term
8x-10<8x
SubIneq
LHS-8x
-10<0
Notice that the variable has been canceled out. However, because -10<0 is always a true statement, any real 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 the entire number line.
Ignacio can unlock the door by giving any number.
b
If we follow a similar thought process, we can find the inequality for Tearrik's riddle. His statement can also be divided into three parts that can be expressed as algebraic expressions.
The expression greater than or equal to can be written as ≥, so the riddle implies a non-stric inequality. Let x represent the unknown number. The first part of the riddle can then be written as 5x-3, and the last part can be expressed as 2x+3(x+2).
The following inequality corresponds to Tearrik's riddle. 5x-3≥2x+3(x+2) Now let's use 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 *
A false statement has been reached. This means there is no number that makes the inequality true and the inequality has no solution.This can be represented with a number line that has no shading.
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Closure
Applying the Properties of Inequalities to Make the Best Choice
In this lesson, we solved inequalities with one variable using the Properties of Inequalities. With the help of these properties, we can determine the best salary option for Ignacio. Recall the options offered to him.
- A base pay of $3000 per month plus 12 % of sales.
- A base pay of $5000 per month plus 5 % of sales.
Now we can find the answers for the following questions.
a
Write an inequality for Ignacio's sales earnings that guarantees Option 1 is better than Option 2 for Ignacio.
b
If Ignacio is sure that he will make at least $35 000 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
Option 1
c
Graph:
Hint
a
Start by writing an algebraic expression for each salary option. 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, let's write each option as an algebraic expression. Let s be the earning from sales. Then, Ignacio will get $3000 and 12 % of s.
Option1 3000+0.12s Using the same logic, Ignacio will get the sum of 5000 and 5 % of s for Option 2. Option2 5000+0.05s To write an inequality, we need to join the two options somehow. In this scenario, since we want Option 1 to be the better one, the expression for the first salary option should be greater than the other expression. Therefore, the inequality is strict and use the symbol >. 3000+0.12s>5000+0.05s
b
Ignacio knows that he can make at least $35 000 on sales. If this value is in the solution set of the inequality, Ignacio should 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, starting with 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>2000/0.07
UseCalc
Use a calculator
s>28 571.42 8571...
RoundInt
Round to nearest integer
s>28 571
This means that Option 1 is better as long as Ignacio makes at least $28 571 worth of sales. Therefore, Ignacio should choose Option 1 because $35 000 is in the solution set.
c
Consider the inequality we found in Part A.
3000+0.12s>5000+0.05s To graph this inequality, we first need to determine its type. Because it has the inequality symbol >, it is a strict inequality. In the previous part, we found its solution set and the boundary point.
| 3000+0.12s>5000+0.05s | |
|---|---|
| Solution Set | Boundary Point |
| s > 28 571 | 28 571 |
Since the inequality is strict, we will place an open circle on the number line on its boundary point 28 571.
Finally, the rest of the solution set needs to be shaded. Because the earnings need to be greater than 28 571, the region on the right side of the boundary point will be shaded.
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Solving and Graphing One Variable Inequalities
Exercise 2.1
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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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["d"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["d<64.17","64.17>d"]}}
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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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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["C"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\dfrac{C}{2\\pi}>9","9<\\dfrac{C}{2\\pi}"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["C"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["C>18\\pi","18\\pi<C"]}}
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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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