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| Student Learning Objectives: |
|---|
|
| | 13 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
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!
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.
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.
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
If x
This property holds for the other types of inequalities.
Identity Property of Addition
Rewrite 0 as z-z
Commutative Property of Addition
- a-b=-(a+b)
Add parentheses
Using the biconditional statement, the last inequality can be rewritten. (y+z)-(x+z)>0 ⇕ x+z
If x
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
If x
This property holds for the other types of inequalities.
Identity Property of Addition
Rewrite 0 as (- z)-(- z)
Commutative Property of Addition
Remove parentheses
- a-b=-(a+b)
Commutative Property of Addition
Add parentheses
From the biconditional statement, the last inequality can be rewritten. (y-z)-(x-z)>0 ⇕ x-z
If x
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.
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.
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
This property holds for the other types of inequalities.
Using these properties, the following conditional statements can be proven.
Each conditional statement will be analyzed separately.
It is given that x
Again, because x
Distribute (- z)
(- a)b = - ab
a-(- b)=a+b
LHS+zy>RHS+zy
Finally, because x
If x
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
This property holds for the other types of inequalities.
Using these properties, the following conditional statements can be proven.
Each case will be analyzed separately.
It is given that x
If x
Again, because x
Put minus sign in numerator
-(b-a)=a-b
Write as a difference of fractions
LHS+y/z>RHS+y/z
Finally, because x
If x
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.
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.
A solution of an inequality is any value of the variable that makes the inequality true. As an example, consider the following inequality. 2x-3< 5 Notice that if 0 is substituted for x in the inequality, the inequality holds true. Therefore, it can be said that 0 is a solution to the given inequality. 2( 0)-3 ? <5 ⇒ - 3 < 5 ✓ However, this is not the only value that makes the inequality true. There are other x-values like 1 and 2 that make it true. The set of all possible values that satisfy an inequality is the solution set of an inequality. The solution set can be determined by applying the Properties of Inequalities to isolate the variable on one side of the inequality.
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.
Therefore, the boundary point is 6 and the solution set corresponds to all real numbers less than 6.
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.
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.
They need to know which of the graphs describes the solution set of the inequality. Let them know which is the correct graph.
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
The boundary point is 3.2, representing the minimum speed that needs to be added to the rocket. This inequality also means that improvements that cause the rocket to speed up greater than or equal to 3.2 kilometers per second will enable the rocket to start its mission.
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.
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.
Which of the graphs describes the solution set of the inequality?
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
LHS-5c>RHS-5c
Subtract term
.LHS /30.>.RHS /30.
Use a calculator
Round to 2 decimal place(s)
The boundary point is about 9.97. This means that as long as Ignacio bought more than 9.97 cryptocurrencies, he can expect to make a profit. Therefore, the minimum he needs to buy is 10.
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.
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.
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
From this inequality, it can be seen that 14 is the boundary point. Recall that m represents the number of mugs. Therefore, m must be a positive integer. Combining these findings, the solution set of the inequality can be written as follows. 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.
| 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.
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.
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Ignacio's Riddle |
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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. |
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Tearrik's Riddle |
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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.
Graph:
Graph:
The expression is less than can be written as <. Therefore, the riddle implies a strict inequality. If x represents the unknown number, then the first part of the riddle can be written as 4x-2. The last part can be expressed as 8x+62.
As a result, Ignacio's riddle is the following inequality. 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, 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.
LHS* 2 < RHS * 2
Distribute 2
LHS-6
Subtract term
LHS-8x
Note that the variable has been canceled out. However, because -10<0 is always a true statement, any number will make the inequality true. Therefore, the solution set of the inequality is all the real numbers. This can be represented on a number line by shading all the number line.
Ignacio can unlock the door by giving any number.
The expression greater than or equal to can be written as ≥. Therefore, the riddle implies a non-stric inequality. Now, let x represent the unknown number. Then the first part of the riddle can be written as 5x-3. 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) 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.
Distribute 3
Add terms
LHS+3≥RHS+3
Add terms
LHS-5x≥RHS-5x
Solving the inequality, a false statement has been reached. This means there is not a number that makes the inequality true. Therefore, the inequality has no solution.This can be represented with a number line that has no shading.
This was a trap all along! Tearrik has no way to escape. Ignacio will be the winner of the escape room.
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.
Now, the answer for the following questions can be found.
First Salary Option 3500+0.12s In a similar way, Ignacio will get the sum of 5000 and 5 % of s. Second Salary Option 5000+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
LHS-3000>RHS-3000
Subtract term
LHS-0.05s>RHS-0.05s
Subtract term
.LHS /0.07.>.RHS /0.07.
Use a calculator
Round to nearest integer
This means that option 1 should be preferred as long as at least $28 571 worth of sales are made. Therefore, Ignacio should choose option 1 because $35 000 is in the solution set.
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 > 28 571 | 28 571 |
Since the inequality is strict, an open circle will be drawn 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.
Consider the following inequality. 3+5x<8-4x Which of the following x-values are solutions to the inequality? x=-2, x=0, x=1
To determine which of the given values are solutions to the inequality, we will evaluate the inequality in each value and simplify. If a true statement is reached, the value is a solution to the inequality. Let's first evaluate the inequality when x= -2.
In a similar fashion, we can determine if x= 0 is a solution to the inequality.
Finally, let's evaluate the inequality when x= 1.
We can now state which of the values are solutions to the inequality. x=-2 and x=0
The perimeter P of a rectangle is the sum of 2 times its length l and 2 times its width w. P=2l+2w The given rectangle has a length of x+5 and a width of x. Let's substitute these expressions into the formula to determine the perimeter of the rectangle.
The perimeter of the rectangle must be greater than 62. That means the inequality is strict and its symbol is >. We can now write the inequality. 4x+10>62
Using the Properties of Inequalities will help us to determine the solution set of this inequality. In this case, we will first apply the Subtraction Property of Inequality.
This means that any value of x greater than 13 will make the perimeter of the rectangle greater than 62.
We will solve the inequality step by step using the Properties of Inequalities. Then we will compare the result obtained in each step with Ramsha's solution to identify the mistake. We will first apply the Addition Property of Inequality to move the constant terms to the right-hand side of the inequality.
Comparing this result with the one Ramsha obtained in the first step, we can conclude that this step has no mistakes. Now, let's apply the Subtraction Property of Inequality to move the variable terms on the left-hand side of the inequality.
Again, comparing this result with Ramsha's result, we see that the Subtraction Property was correctly applied. Finally, let's apply the Division Property of Inequality to isolate the variable. Since we need to divide by a negative number, -4, the inequality symbol must be reversed.
Note that our result is different from the one Ramsha got. The symbols are the opposite. This means that Ramsha forgot to change the inequality symbol when dividing by a negative number. Therefore, the wrong step is Step 3.
In Part A we already solved the inequality and got x > -2 as the correct solution.
Consider the following values. 7, 5, 4, 3, 1 Which value of x does not meet the condition 3x+1>7?
To identify which of the values meet the condition of 3x+1>7, we will solve the inequality using the Properties of Inequalities.
The solution set of the given inequality is given by all values greater than 2. Solution Set: {x|x>2} Any value that is not greater than 2 will not meet the condition 3x+1>7. From the given values, only 1 is not greater than 2. Therefore, the value of x which does not meet the condition is x=1.
Consider the given inequality. 5x+3<7-2x We will apply the Properties of Inequalities to isolate the variable on one side of the inequality. To do so, we will begin by applying the Addition Property of Inequality.
The variable term is already on one side of the inequality. The Division Property of Inequality will be applied now to isolate the variable. Consider that if an inequality is divided by a negative number, the inequality symbol needs to be reversed.
Consider the given inequality. 2(x-6)+2≥4(2x+5) We will begin by simplifying both sides of the inequality.
We can now apply the Properties of Inequality. Let's first apply the Subtraction Property of Inequality.
The variable term is already on one side of the inequality. We will divide both sides of the inequality by 6 to find its solution set.
Consider the given inequality. y/-6≥6 To solve this inequality, we need to apply the Multiplication Property of Inequality. Consider that when multiplying an inequality by a negative number, the inequality symbol needs to be reversed.
Consider the given inequality. -2x≤ 22-x/6 In a similar fashion, we will use the Properties of Inequalities to find the solution set of the given inequality. In this case, we will begin by applying the Multiplication Property of Inequality.
The variable term is already on the left-hand side of the inequality. We will apply the Division Property of Inequality to isolate the variable. Since we will divide both sides of the inequality by -11, the inequality symbol needs to be reversed.