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| Student Learning Objectives: |
|---|
|
| | 10 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
From the school yard to the kitchen, three states of water are observable at a moments notice: solid, liquid, and gas.
It is known that, at atmospheric pressure, water freezes at 32^(∘) F and vaporizes at 212^(∘) F.
An inequality that results from combining two inequalities by using the the word and
or the word or
, is called a compound inequality. The applet shows compound inequalities and their solution sets. Examine the solution set for the indicated inequality and explore how it changes for different compound inequalities.
An absolute value inequality is an inequality that involves the absolute value of an expression containing a variable. As with other inequalities, absolute value inequalities can be strict or non-strict.
| Strict Absolute Value Inequalities | Non-Strict Absolute Value Inequalities | ||
|---|---|---|---|
| |x+2| > 5 | |x+7| < 5 | |2x| ≥ 10 | |x-2| ≤ 4 |
If a≥ 0, an absolute value inequality of the form |x|< a can be seen as the set of all numbers that are greater than - a and less than a. Similarly, an absolute value inequality of the form |x|≤ a can be seen as the set of all numbers that are greater than or equal to - a and less than or equal to a. c|c |x|< a & |x|≤ a ⇕ & ⇕ - a < x < a & - a ≤ x ≤ a Likewise, if a≥ 0, an absolute value inequality of the form |x|> a can be seen as the set of all numbers that are less than - a or greater than a. Similarly, an absolute value inequality of the form |x|≥ a can be seen as the set of all numbers that are less than or equal to - a or greater than or equal to a. c|c |x|> a & |x|≥ a ⇕ & ⇕ x < - a or x > a & x≤ - a or x≥ a As with other inequalities, absolute value inequalities can be represented by an interval on a number line. Open points at the ends of the interval represent strict inequalities where the corresponding values are not included in the interval. Conversely, closed points represent non-strict inequalities and the corresponding values are included in the interval.
Using the definition of absolute value, an absolute value inequality can be written as a compound inequality. Therefore, solving absolute value inequalities is quite similar to solving compound inequalities. To detail the reasoning, the following inequality will be solved. 3 |x-7| +2 > 20 There are four steps to follow.
Here, the number on the other side of the absolute value is positive. However, it should be noted that if that number were negative, then the inequality either has no solution or all real numbers as a solution depending on the inequality symbol. Example Inequality & Solution Set [-1em] |x-7|< - 5 & No Solution |x-7| > - 5 & All Real Numbers Since the absolute value of an expression is always non-negative, |x-7| should be greater than or equal to 0. Therefore, all real numbers satisfy |x-7|> - 5, and there is no x-value which satisfies |x-7|< - 5.
or,the combination of the solution sets is also written with that word. x < 1 or x > 13
Izabella goes on a tour of a chocolate factory. They aim to produce chocolate bars weighing 93 grams. A machine at the factory weighs chocolates chosen at random. The bars must not deviate from the predetermined weight by more than 5 grams. Otherwise, the machine has to send them back.
88 ≤ w At the same time, it must weigh less than or equal to 93+5=98 grams. w ≤ 98 Consequently, the combination of these individual inequalities will result in a compound inequality describing the range of acceptable weights. 88 ≤ w and w ≤ 98 ⇕ 88 ≤ w ≤ 98
The other inequality represents all the points to the left of 98. Again, the inequality sign is non-strict, so 98 is included.
Because the inequalities are combined using word and,
the solution set of the resulting compound inequality is equal to the union of the solution sets of the individual inequalities. Therefore, the graph of the acceptable weight range is the combination of the graphs.
|w-93| ≤ 5 Since this absolute value inequality can be rewritten as the compound inequality obtained in the previous steps, both have the same set of solutions.
Izabella, looking around the chocolate factory, discovers a room full of chocolate fondue fountains on sale for special events! The prices are high and they vary significantly. She decides to make a list of the prices to see which is a fair price.
Substitute values
Since Izabella thinks a fair amount is within $100 of the average price of $447, the difference between the amount x Izabella thinks is fair and the average price should be less than $100. Therefore, the absolute value of the difference between x and the average price is less than 100. Absolute Value Inequality [0.6em] |x-447| < 100 By solving this absolute value inequality, the chocolate fondues satisfying this condition can be determined. To do so, first rewrite it as a compound inequality and then solve for x. |x-447| < 100 ⇓ - 100 < x-447 < 100 ⇓ 347 < x < 547 The chocolate fondues whose prices are within this range are fair prices.
As seen, 4 prices meet Izabella's condition. $ 450, $ 476, $ 358, and $480
|x-447| ≤ 20 By solving the inequality, the fondue prices satisfying this condition can be determined. |x-447| ≤ 20 ⇓ - 20 ≤ x-447 ≤ 20 ⇓ 427 ≤ x ≤ 467 The fondue prices are within this range are those which Izabella would consider to be priced fairly.
There is only one price that meets Izabella's condition. $ 450
Izabella learns that the factory has future plans to develop packaging that can withstand the harshest conditions, including Mars! The temperature of the surface of Mars reaches its highest value at the equator where it is less than 36^(∘) C. Mars reaches its lowest temperature at the poles where it is always greater than - 144^(∘) C.
Write an absolute value inequality for the range of possible temperature values t on Mars.
To find d, half of the difference between the endpoints will be calculated.
The distance d to the midpoint from the endpoints is 90. Therefore, the midpoint is the number - 144+ 90 = - 54.
The points that are within 90 units of the midpoint represent the range can be written as the following absolute inequality. Note that since the endpoints are not included, the inequality should be strict. |t-( - 54)| < 90 ⇕ |t + 54| < 90 Now that Izabella has this information, she is even more impressed with the chocolate factory!
At this special chocolate factory the average salary for a Chocolatier is a whopping $ 45 700.
As a company policy, a new Chocolatier's actual salary can only differ from the company average by less than $1250.
Range: 44 450 < s < 49 950
Absolute Value Inequality [0.6em] |s-45 700| < 1250 By solving this absolute value inequality, the range for the possible salaries can be found. To do so, the inequality will be rewritten. |s-45 700| < 1250 ⇓ - 1250 < s-45 700 < 1250 ⇓ 44 450 < s < 46 950 The solution set of the inequality is the set of values between 44 450 and 46 950. Since the inequality sign is strict, those numbers are not included.
The values that are more than 1250 units away from the average salary in the number line, need to be represented by an absolute value inequality.
These values can be written as the following absolute value inequality. |s-45 700| ≥ 1250
Analyze the given graph and determine the corresponding absolute value inequality.
In this lesson, the relationship between absolute value inequalities and compound inequalities has been explained using real-world examples. Considering those examples, the challenge presented at the beginning can now be solved seamlessly.
It is known that, at atmospheric pressure, water freezes at 32^(∘) F and vaporizes at 212^(∘) F.
Solid: & t < 32 Gas: & t > 212 The values 32^(∘) F and 212^(∘) F are not included because at these temperatures water starts to change its physical state and the liquid form of water can be present in both states of the transition. Under these conditions, the following compound inequality shows the temperatures in which water is not a liquid. t<32 or t>212
The other inequality represents all of the points to the right of 212. Again, the inequality sign is strict, so 212 is not included.
Since the inequalities were combined with the word or,
the solution set of the resulting compound inequality is the union of the solutions sets of the two individual inequalities. Therefore, the graph of the range is the combination of the above graphs.
212-32/2= 180/2 ⇒ 180/2=90 The midpoint is 90 units away from the endpoints. Therefore, the midpoint is 32 + 90 = 122.
The points that are further away in units than the calculated distance of 90 represent the range, which can be written as the following absolute inequality. |t-122|>90
Consider the compound inequality. |x-7| < 5 and |x+3|> 7 Which of the following represents the solution set of the inequality?
We have been given a compound inequality that is composed of two absolute value inequalities. Let's start by rewriting and graphing each of the absolute value inequalities.
The first absolute value inequality means that the distance between x and 7 is less than 5, which can be rewritten as an and
compound inequality.
|x-7|< 5
⇓
- 5 < x-7 < 5
Let's solve this inequality!
We can graph this solution set on a number line.
The second absolute value inequality represents the distance between x and - 3 is greater than 7. In other words, it represents the set of all numbers that are less than - 7 or greater than 7. This can be rewritten as an or
compound inequality.
|x+3|> 7
⇓
x+3<- 7 or x+3> 7
Let's solve each individual inequality!
c|c
Inequality I & Inequality II [0.7em]
x+3 < - 7 & x+3 > 7
⇓ & ⇓
x < - 10 & x > 4
We can graph this solution on a number line.
To find the solution set of the compound inequality, we can graph both solution sets on the same number line.
Since the compound inequality is an and
compound inequality, the solution set to the compound inequality is the intersection of the solution sets of the absolute value inequalities.
The solution set is the overlapping interval between 4 and 12.
| Compound Inequality | |x-7| < 5 and |x+3|> 7 |
|---|---|
| Solution Set | 4 < x < 12 |