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| | 15 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Jordan is a member of the drama club at her school. The club plans to stage the play The Little Frog in Town this weekend. They charge $6 per ticket.
Not all inequalities are expressed in the form x< a, x > a, x ≤ a, or x ≥ a. Yet, through inverse operations and the Properties of Inequalities, any inequality can be simplified to one of the mentioned forms. Consider the Addition and Subtraction Properties of Inequalities.
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
Identity Property of Addition
Rewrite 0 as z-z
Commutative Property of Addition
- a-b=-(a+b)
Add parentheses
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
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
If x
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
The case when x
Using these properties, the following conditional statements can be proven.
Each conditional statement will be analyzed separately.
It is given that x
Distribute (- z)
(- a)b = - ab
a-(- b)=a+b
LHS+zy>RHS+zy
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
The case when x
Using these properties, the following conditional statements can be proven.
Each case will be analyzed separately.
It is given that x
If x
Put minus sign in numerator
-(b-a)=a-b
Write as a difference of fractions
LHS+y/z>RHS+y/z
If x
In the applet, determine the property used to isolate the variable on one side of the given inequality as shown.
The play The Little Frog in Town received rave reviews from the audience after its first showing. The club decides to put on this play again with one major difference — they will hold it in the greatest theater hall their city has to offer! Now, the drama club needs to know the number of seats in the hall to print new tickets.
The hall consists of two floors, the ground floor and the balcony. The number of seats s on the ground floor is defined by the following inequality. 4s-132 ≤ s+174 The number of seats b in the balcony is defined by another inequality. 143-3b > b-17
ccc Ground Floor & and & Balcony s ≤ 102 & & b < 40 According to these inequalities, there are at most 102 seats on the ground floor and 39 seats on the balcony. Now, add these number of seats to find the maximum number of seats in the theater hall. 102+39=141 The maximum number of seats in the theater hall is 141.
After watching excellent theater performances put on by the drama club, so many students want to join it. Several characteristics are highly sought after for new recruits.
For instance, at the moment, the drama club does not have many tall members. For this recruiting year, they have defined a height to consider for new recruits h with the following inequality. -2h+30 ≤ h-315 More importantly, the new recruit needs to get a great score s in their Literature & Language class. The score is defined with the following inequality. 5(s-14) > 245
Divide by -3 and flip inequality sign
Calculate quotient
.LHS /5.>.RHS /5.
Cancel out common factors
Calculate quotient
In the applet, there are given various different inequalities. Select the correct solution set for the given 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 <. x + 2 < 8_(Strict)
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.
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.
The drama club makes a survey of several audiences about the plays presented during the school year. Then they discussed these surveys' results at the end-of-year meeting.
This year all plays were at least 100 minutes long. Since the audiences found the duration of the plays a bit too long according to the survey results, the club takes a decision to keep the next year's plays less than 90 minutes in length.
d < 90 Notice that it can also be written by using the phrasegreater than. In other words, saying d is less than 90 is the same as saying 90 is greater than the duration of the plays d. d < 90 ⇕ 90 > d
d < 90 Since the inequality is strict, 90 is not included in the solution set. This means that 90 will be shown with an open circle in the graph. This already eliminates options II and III. Also, notice that the solution set needs to include real numbers, not just integers, because time is a continuous quantity. Solution Set {d | d is a real number less than 90} The option IV does not make sense for this solution set because it shows only integers. However, Graph I shows exactly the real numbers less than 90.
This means that the answer is Graph I.
.LHS /6.≤.RHS /6.
Calculate quotient
Round to 2 decimal place(s)
Izabella uses the subway to commute to school. A ticket for a single ride costs a student $1.15. The subway company also offers a monthly pass for $45 which gives access to an unlimited number of rides.
We want to write an inequality that represents how many times Izabella can ride the subway in a month so that a monthly pass is a better deal. We know that a single subway ride costs $ 1.15 and a monthly pass costs $ 45.
| Subway Rides Costs | |
|---|---|
| Single-ride Ticket | $ 1.15 |
| Monthly Pass | $ 45 |
Let x be the number of times Izabella rides the subway in a month. We can find the price of x subway rides on single-ride tickets by multiplying x by the price of a single pass. 1.15x Izabella can also buy a monthly pass and pay $ 45 for x rides. If a monthly pass is a better deal, then it is cheaper than 1.15x. This means that 1.15x is more than 45. Let's write it as an inequality. 1.15x > 45
It is a given that Izabella rides a subway about 50 times per month. In Part A we found the inequality that represents how many times Izabella can ride the subway so that a monthly pass is a better deal than buying single-ride tickets.
1.15x > 45
In the inequality the variable x is the number of times Izabella rides the subway in a month. We will substitute 50 for x and check whether the inequality produces a true statement.
The inequality produces a true statement for x = 50. This means that buying a monthly pass is a better deal than buying single-ride tickets.
Izabella's mom gives her $55 to go shopping for a dance. She decides buy a dress. It costs $32.
Izabella can spend up to $ 55 and she wants to buy a dress that costs $ 32. Let x be the remaining amount of money that she can spend after buying the dress. We will write and solve an inequality to find the possible values of x. The total amount of money that she can spend is the sum of the dress price and x. 32 + x We know that she can spend up to $ 55. This means that the total amount of money that she can spend is less than or equal to 55. Let's write it as a non-strict inequality. 32 + x ≤ 55 Next, we will subtract 32 from each side of the inequality to isolate the variable x by using the Subtraction Property of Inequality.
This means that Izabella can spend at most $23. The remaining amount of money after buying the dress can be shown by the following inequalities. 32 + x ≤ 55 ⇕ x ≤ 23
The dress that costs $ 32 is on sale for 15 %. We will find the new price of the dress and rewrite the inequality from Part A. First, let's find the amount of discount by calculating 15 % of 32.
The new price of the dress is the original price minus the amount of the discount. 32- 4.8 = 27.2 We found that the price of the dress on sale is $ 27.20. In Part A we wrote the inequality when the price of the dress is $ 32. The inequality can be used to find the amount of money that Izabella can spend, x. Let's consider the inequality from Part A. 32 + x ≤ 55 The new price of the shirt is $ 27.20. We will write the same inequality but instead of 32, we will write 27.2. 27.2 + x ≤ 55 Notice that we can rearrange the inequality by isolating the variable to solve it. Let's subtract 27.2 from both sides of the inequality.
This means that Izabella can spend at most $27.80 after buying the dress on sale. 27.2 + x ≤ 55 ⇕ x ≤ 27.8
Students in a science class are divided into 4 equal groups. Let n be the number of students in the science class. We can find the number of students in each group by dividing the total number of students by the number of groups. n/4 We know that in each group there are at least 5 students. This means that n4 should be greater than or equal to 5. Let's write it as an inequality. n/4 ≥ 5 We will solve the inequality to find the possible values of n. Let's multiply both sides of the inequality by 4 by using the Multiplication Property of Inequality.
This means that the number of students should be greater than equal to 20. Notice that the number of students should be a natural number. Let's compare the given numbers with 20. 16 < 20 * 18 < 20 * 20 = 20 ✓ 24 ≥ 20 ✓ 30 ≥ 20 ✓ It was also said that the groups are equal which means they should include equal number of students. Since there are 4 groups, we should consider the numbers that are divisible by 4. With this in mind, let's consider the numbers greater than equal to 20 once again. 20 ÷ 4 =5 ✓ 24 ÷ 4=6 ✓ 30 ÷ 4=7.5 * Since the number of students should be a natural number, there cannot be 7.5 students in a group. The possible numbers are 20 and 24.