13. Solving and Graphing One Variable Absolute Value Inequalities
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Chapter 1
13.
Solving and Graphing One Variable Absolute Value Inequalities
Understanding absolute value inequalities is crucial. These inequalities involve the absolute value of an expression containing a variable. Through real-world examples, such as determining the acceptable weight range for chocolate bars or understanding the temperature variations on Mars, one can grasp the practical applications of these inequalities. For instance, a chocolate factory aims to produce bars weighing a specific amount, and any deviation beyond a set limit is unacceptable. Similarly, the temperature on Mars has its highs and lows, and understanding these ranges is essential for future explorations. By mastering absolute value inequalities, one can tackle complex problems with ease and precision.
Show less Show more expand_more| Student Learning Objectives: |
|---|
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| | 10 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving and Graphing One Variable Absolute Value Inequalities
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What can ensure a quality approach when analyzing elements such as water or soil? How about when making chocolate? Predetermined ranges for specific characteristics, that is. For instance, it is possible to use inequalities that involve absolute values to ensure the quality of a chocolate bar or in understanding elements of nature. Here, one-variable absolute value inequalities will be solved, and their solution sets will be drawn.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Representing Water Temperature in Non-liquid States With Absolute Value Inequality
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.
a
Write the range of temperatures in which water is not liquid.
b
Graph the given range of when water freezes and vaporizes.
c
Write the absolute value inequality that describes this situation.
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Explore
From Compound Inequalities to Absolute Value Inequalities
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.
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Discussion
Absolute Value 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.
Method
Solving an Absolute Value Inequality
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.
1
Isolate the Absolute Value Expression
expand_more To be able to rewrite the given inequality as a compound inequality, the absolute value expression should be isolated first. To do so, the Properties of Inequalities and inverse operations will be used.
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.
2
Rewrite It as a Compound Inequality
expand_more Now, the absolute value inequality |x-7|>6 will be rewritten. It represents the set of all numbers that are less than - 6 or greater than 6.
|x-7| > 6 ⇓ x-7 < - 6 or x-7 > 6
3
Solve the Individual Inequalities
expand_more Next, the individual inequalities will be solved by adding 7 to both sides of the inequalities.
c|c
Inequality I & Inequality II [0.7em]
x-7 < - 6 & x-7 > 6 ⇓ & ⇓ x < 1 & x > 13
The solution set of Inequality I contains all real numbers less than 1. The solutions to Inequality II are all real numbers greater than 13.
4
Combine the Solution Sets
expand_more The combination of the solution sets for the individual inequalities is the solution set of the given absolute value inequality. Since the derived compound inequality was combined with the word
or,the combination of the solution sets is also written with that word. x < 1 or x > 13
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Example
Representing Acceptable Chocolate Weights Using an Absolute Value Inequality
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.
a
Write a range of allowable weights w for a chocolate bar using a compound inequality.
b
Graph the range on a number line.
c
Write an absolute value inequality that describes this situation.
Answer
a
88 ≤ w ≤ 98
b 
Graph:
c
|w-93| ≤ 5
Hint
a
Determine the maximum and minimum acceptable weights.
b
Should the solution set contain the maximum and minimum w-values?
c
Which inequality symbol should be used?
Solution
a
For a chocolate bar to be sent back, the difference between the weight of the chocolate bar and the predetermined weight is more than 5 grams. Therefore, an acceptable weight w for a chocolate bar, not to be sent back, must be greater than or equal to 93-5=88 grams.
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
b
The inequality 88≤ w is the set of values which are greater than or equal to 88. Since the inequality sign is non-strict, 88 itself is included.
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.
c
Any chocolate bar that varies from the predetermined weight by more than 5 grams is sent back. Therefore, the absolute value of the difference between the weight of a bar w and the predetermined weight should be less than or equal to 5. This can be expressed as follows.
|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.
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Example
Writing an Absolute Value Inequality to Define a Range of Prices
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.
a
To define a fair price, Izabella decides a fondue with a price within $100 of the average price of all the fondues is fair. Write an absolute value inequality describing the situation. Which prices meet this condition?
b
Izabella decides to narrow the range. She thinks that a fondue that is no more than 20 above the average price is what is really fair. Write an absolute value inequality for this case and determine the prices that meet this condition.
Answer
a
Absolute Value Inequality: |x-447| < 100
Prices: $ 358, $ 450, $ 476, and $ 480
b
Absolute Value Inequality: |x-447| ≤ 20
Price: $ 450
Hint
a
The average price of the chocolate fondues is the sum of the prices divided by the number of prices. The absolute value of the difference between the amount of money Izabella thinks is fair and the average price should be less than 100.
b
The absolute value of the difference between the amount of money Izaballa thinks is fair and the average price should be less than or equal to 20.
Solution
a
To find the average price p of the chocolate fondues, the sum of the prices will be divide by the number of prices, 8.
p = Sum of Values/Number of Values
SubstituteValues
Substitute values
p = 450 + 476+560+ 327+358+325+600+480/8
p = 447
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
b
In the previous part, the average price of the fondue's was found to be $447. Now, the range is narrowed to $20. The absolute value of the difference between x and 447 should be less than or equal to 20.
|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
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Example
Representing the Temperature of Mars With an Absolute Value Inequality
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.
External credits: ESA & MPS for OSIRIS Team MPS/UPD/LAM/IAA/RSSD/INTA/UPM/DASP/IDA
Write an absolute value inequality for the range of possible temperature values t on Mars.
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Hint
Start by writing a compound inequality for the possible temperature values t on Mars.
Solution
From the given information, the temperature on Mars is always less than 36^(∘) C but greater than - 144^(∘) C. At the equator: & t < 36 At the poles: & t > - 144
These individual inequalities can be combined to form an equivalent compound inequality that describe the possible temperatures on Mars.
- 144 < t and t < 36 ⇕ - 144 < t < 36
To write this compound inequality as an absolute value inequality, the midpoint between - 144 and 36 on the number line should be found. Let d be the distance to the midpoint.
To find d, half of the difference between the endpoints will be calculated.
d = 36- (- 144)/2
d = 90
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!
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Example
Representing a Salary Range With an Absolute Value Inequality
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.
a
Write and solve an absolute value inequality to find a range for the possible salaries s for a new Chocolatier. Draw the range on a number line.
b
Write an absolute value inequality for the salaries that are not offered to new Chocolatiers.
Answer
a
Absolute Value Inequality: |s-45 700|<1250
Range: 44 450 < s < 49 950
b
Absolute Value Inequality: |s-45 700|≥ 1250
Hint
a
The absolute value of the difference between the salary for a new Chocolatier and the average salary of any Chocolatier is less than 1250.
b
Use the graph drawn in the part A.
Solution
a
This special chocolate factory pays a new Chocolatier's salary within $1250 of the company's average salary of $ 45 700. This means that the absolute value of the difference between the salary s for a new employee and the average salary of any Chocolatier is less than 1250.
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.
b
Consider the graph of the absolute value inequality found in Part A.
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
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Pop Quiz
Determining an Absolute Value Inequality From Its Graph
Analyze the given graph and determine the corresponding absolute value inequality.
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Closure
Representing Water Temperature in Non-liquid States With 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.
a
Write the range of temperatures in which water is not liquid.
b
Graph this range.
c
Write the absolute value inequality that describes this situation.
Answer
a
t< 32 or t > 212
b 
Graph:
c
|t-122|>90
Hint
a
Determine the values of temperature that water freezes or vaporizes.
b
Does the solution set contain the numbers 32 and 212?
c
Find the middle point between 32 and 212 on the number line.
Solution
a
Water is not liquid when it is in frozen form nor when it is vapor form. Noting the given information, it is known that, at atmospheric pressure, water is not liquid for temperatures less than 32^(∘) F or greater than 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
b
The compound inequality is the combination of two inequalities. The inequality t<32 shows the set of values less than 32. Since the inequality sign is strict, the value 32 is not included.
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.
c
To write the absolute value equation, the midpoint between 32 and 212 on the number line should be found. To do so, half the difference between these numbers will be calculated.
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
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Solving and Graphing One Variable Absolute Value Inequalities
Exercise 2.1
An auto parts manufacturer produces gaskets for cars. They throw out gaskets that do not weigh within 0.06 pound of the mean weight of a batch.
External credits: Sonett72
The weights (in pounds) of the gaskets in a batch are 0.57, 0.64, 0.65, 0.52, and 0.62. Which gasket(s) should be thrown out?
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The gaskets are only thrown out if they are not within 0.06 pounds of the mean weight of the batch. The first thing we need to do is to calculate the mean weight of the batch so that we can know our central point of comparison.
The mean weight of the batch is 0.6 pounds. We need the individual gasket weights w to be less than 0.06 pounds away from 0.6. We can solve for this range of values by writing an absolute value inequality.
|w-0.6|<0.06
Since we need the distance from 0.6 to be less than 0.06, we will need to write an and
compound inequality.
w-0.6<0.06 and w-0.6>-0.06
We can solve these cases separately and then combine the results. We will begin with the first case.
Let's solve the second case.
With this information we can write the solution set for this compound inequality. w<0.66 and w>0.54 The acceptable range of weights for the gaskets are those that are less than 0.66 pounds and greater than 0.54 pounds. Therefore, the gasket that needs to be thrown out is the one weighing 0.52 pounds.
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Solution
The difference between the areas of the figures is less than 3.
What is the possible integer value of x?
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Let A_T be the area of the triangle and A_R be the area of the rectangle. We are told that the difference between the areas of the figures is less than 3. We can write this as an absolute value inequality. |A_T-A_R|< 3 We will now find the area of each figure.
Area of Triangle
The area of a triangle with base b and height h can be found using the following formula. A_T=1/2bh We can see from the given diagram that the base of the triangle is 6 and its height is x+4. Let's substitute these values into the formula for the area of a triangle.
Area of Rectangle
The area of a rectangle with length l and width w can be found using the following formula. A_R=l w We can see from the given diagram that the length of the rectangle is 4 and its width is 3. Let's substitute these values into the formula for the area of a rectangle.
Absolute Value Inequality
We can now substitute these expressions into our absolute value inequality. | 3x+12-12 |< 3 ⇓ |3x| < 3 To solve this inequality for x, we will first rewrite it as a compound inequality. |3x| < 3 ⇓ 3x > - 3 and 3x < 3 We will now solve the individual inequalities by dividing both sides by 3. c|c Inequality I & Inequality II [0.7em] 3x > - 3 & 3x < 3 ⇓ & ⇓ x > - 1 & x < 1 We can combine the solution sets of the individual inequalities, which gives the solution set of the absolute value inequality. - 1 < x < 1 In this interval, 0 is the only possible integer value for x.
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Solution
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{"type":"choice","form":{"alts":["True","False"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
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To verify if a satisfies the two given inequalities, we rewrite the absolute value inequality as a compound inequality. Since the absolute value is less than or equal to 10, we use the word and
when writing the compound inequality.
|x+9|≤ 10
⇓
x+9≥- 10 and x+9≤ 10
Notice that one of these inequalities is the same as the given inequality. This means that a will also be a solution to x+9≥- 10. Therefore, the statement is true.
We will again start by rewriting the absolute value inequality as a compound inequality. In this case, we can rewrite the absolute value inequality as an or
compound inequality because the absolute value is greater than or equal to.
|x+9|≥ 10
⇓
x+9≥ 10 or x+9≤- 10
Notice that one of the inequalities is the opposite of the given inequality. This means that an arbitrary number a cannot satisfy both inequalities. We will verify this by supposing that x= - 21. In the compound inequality, this value satisfies the second inequality.
- 21+9≤- 10 ⇒ - 12 ≤- 10 ✓
However, it does not satisfy the other inequality.
-21+9 ? ≥- 10 ⇒ -12 ≱-10 *
Therefore, the statement is false.
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Solution
Izabella claims that the solution of |n|>0 is all real numbers. Is she correct?
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We know that any non-zero value, positive or negative, substituted for n will give a positive result. |a|=a and |- a|=a Therefore, the absolute value of any non-zero number is greater than 0. However, substituting 0 for n makes the given inequality false because a number cannot be greater than itself. 0 ≯ 0 As a result, the solution of the given inequality is all real numbers except 0. We conclude that Izabella's claim is false.
For a better understanding of the inequality, let's represent the solution set of the absolute value inequality |n|>0 on a number line. Since 0 is the only number that does not satisfy the inequality, it should be open. We need to show the rest of the number line.
If the inequality symbol was ≥, the solution set would be the whole number line, without any open circle.
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Solution
Solving and Graphing One Variable Absolute Value Inequalities
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