9. Solving Absolute Value Equations
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Chapter 1
9.
Solving Absolute Value Equations
Absolute value equations are mathematical expressions that involve the absolute value of a variable. This lesson explores the methods for solving such equations, including algebraic and graphical approaches. It also delves into the properties of equality that are essential for isolating the absolute value expression. Real-world applications are highlighted, such as predicting distances and modeling battery performance. For example, the distance a train is from a village can be modeled using absolute value equations to determine when a smartphone will regain signal. The lesson provides a comprehensive guide for understanding and applying these equations in various scenarios.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Absolute Value Equations
Slide of 10
Mathematical models, at times, can predict irrelevant values when describing real-world scenarios. For example, the calculation of a length only makes sense to obtain a non-negative result. Thankfully, this lesson will show how situations that require a positive result can be modeled using the absolute value. Additionally, equations that involve absolute value expressions will be explored.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
- Variable
- Expression
- Number Line
- Solution of an Equation
- Opposite of a Number
- Properties of Equality
- Inverse Operations
- Solving Multi-Step Equations
Here are a few practice exercises before getting started with this lesson.
a Which of the following expressions represents the solution to the equation 3x-4=14?
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b How many units away is -5 from 0 on the number line?
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c Which two values are at the same distance from 2 on the number line?
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Challenge
Modeling the Performance of a Battery
A university is developing an eco-friendly battery for tablets called Flora, which uses no harmful chemicals for the environment. After running some tests, the following number line describes the results. Fully charged, the point at 10 indicates the average amount of hours a Flora battery lasts. The points at 8 and 12 indicate the minimum and maximum performance times, respectively.
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Explore
Why and How to Define an Always Positive Quantity
Think of a mathematical model that needs to predict a strictly positive quantity. Ever wonder exactly how many days until the end of school, but are only given the calendar date? Well, consider a formula that counts the number of days. A date is entered into the formula and the prediction of 20 days away from today is made.
Then, another date is entered, but this time the formula says that it is - 30 days away!
Considering the given information about the formula, try to answer the following questions.
- Since days cannot be negative, what can the minus sign mean here?
- How many days away is this date from today?
- If the formula predicted a value of - x for a specific date, how many days away would it be from today?
- Is there a reasonable way to assign a positive value to every real number in general?
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Discussion
Absolute Value
The absolute value of a number a is the distance between a and 0 on the number line. It is denoted as |a| and it is always a non-negative value.
Absolute Value Properties
There are several properties and identities that are useful when simplifying expressions or solving equations dealing with absolute values. For any two real numbers a and b, the following relationships and identities hold true.
| Property | Algebraic Representation |
|---|---|
| Non-negativity | |a| ≥ 0 |
| Symmetry | | - a| = |a| |
| Idempotence | ||a|| = |a| |
| Positive-definiteness | |a| = 0 ⇔ a = 0 |
| Identity of Indiscernibles | |a − b| = 0 ⇔ a = b |
| Multiplicativity | |ab| = |a| * |b| |
| Preservation of Division | |a/b|=|a|/|b| if b ≠ 0 |
| Subadditivity | |a + b| ≤ |a| + |b| |
| Triangle Inequality | |a − b| ≤ |a − c| + |c − b| |
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Pop Quiz
Simplifying Absolute Value Expressions
Practice simplifying absolute value expressions by using the following applet.
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Discussion
Absolute Value Equations
An absolute value equation is an equation that involves the absolute value of a variable expression. | 2x-8 |=5 Equations of the form |x|=a, where a is a real number greater than zero, can be solved by looking for the numbers x whose distance from 0 in the number line equals a. For example, the solutions of |x|=4 are all values of x that are 4 units away from 0.
Since there are two points on the number line that fulfill this requirement, there are two solutions to the equation |x|=4, namely x=4 and x=- 4. However, solving an absolute value equation, in general, might require a more elaborate and structured approach.
Number of Solutions
Simple absolute value equations of the form |x|=a, can have no, one, or two solutions, depending on the value of a. However, more complex absolute value equations may have more than two solutions.
| Equation | Number of Solutions | Solution(s) |
|---|---|---|
| |x|=- 4 | Zero | No solution |
| |x|=0 | One | 0 |
| |x|=4 | Two | - 4, 4 |
| |x^2-4 |=2 | Four | - sqrt(2), sqrt(2), - sqrt(6), sqrt(6) |
Method
Solving Absolute Value Equations Algebraically
An absolute value equation can be solved algebraically by first isolating the absolute value expression. Then, consider the two possible cases for the argument inside the absolute value: one where it is positive and one where it is negative. These lead to two separate linear equations, which can be solved independently. The following example will illustrate this process. |2x+4|-16=0 To solve an absolute value equation, there are four steps to follow.
1
Isolate the Absolute Value Expression
expand_more Use the Properties of Equality and inverse operations to isolate the absolute value expression on one side of the equation.
2
Rewrite the Absolute Value Equation as Two Linear Equations
expand_more The absolute value function |x| gives the distance from zero to x, making it positive or zero. For an equation |ax+b|=k, where k≥0, this implies two possible cases.
ax+b=k or ax+b=- k
Applying this to the absolute value equation rewritten in Step 1 creates two linear equations.
2x+4=16 or 2x+4=-16
3
Solve Each Linear Equation Separately
expand_more Solve each linear equation found in Step 2 separately.
| 2x+4=16 | 2x+4=-16 |
|---|---|
| 2x=12 | 2x=-20 |
| x=12/2 | x=-20/2 |
| x=6 | x=-10 |
4
Combine the Solutions
expand_more The solutions derived from the linear equations satisfy the original absolute value equation, making them solutions to the absolute value equation. Since x=6 and x=-10 are solutions for each linear equation respectively, these are also solutions to the equation |2x+4|-16=0
Equation: &|2x+4|-16=0 Solutions:& x=6 or x=-10
Extra
Negative or Zero Constant TermsAfter writing the absolute value equation in the form |ax+b|=k, where a, b, and k are constants, check the value of k.
- If k<0, there is no solution. The absolute value function always returns a non-negative number, making it impossible for |ax+b| to be negative.
- If k=0, the equation simplifies to |ax+b|=0, which has a single solution given by the linear equation ax+b=0.
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Pop Quiz
Practice Solving Absolute Value Equations
Practice solving absolute value equations by using the following applet. Indicate which number line represents the solution set of the given equation.
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Example
Using Absolute Value Equations to Model Real Life Situations
Davontay is on a train trip to a school of magic. His smartphone is powered by an eco-friendly battery, but still a young product, it easily loses reception. A vendor notices and gives him hope. She exclaims, "Don't worry, lad. When the train is 60 kilometers near the next wizard village, the phone will get signal. It will last until the train is 60 kilometers past that village."
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Hint
The current distance from the train to the village is 150 kilometers, and for each passing minute its distance to the next village is reduced by 3 kilometers. When will the trains distance be equal to 60 kilometers?
Solution
This scenario will be solved using two methods. First, by using a number line, and then, it will be solved through setting up and solving an absolute value equation.
Solving the Example by Using a Number Line
It is helpful to summarize the important information in a diagram. Recall that the train is originally 150 kilometers away from the next wizard village, and the smartphone will only get a signal when the distance of the train from the village is equal to 60 kilometers.
From the diagram, it can be seen that the train needs to cover a distance of 150-60=90 kilometers for the smartphone to recover its signal. By using the speed formula, the time of travel for that distance can be calculated.
Therefore, the phone will recover its signal after 30 minutes. Now, to determine how long it will keep connected, the time the train will take to leave the 60 kilometers proximity from the wizard village should be found.
As can be seen, the train needs to cover a distance of 150+60=210 kilometers for the smartphone to lose connection again. Now the formula for speed will be used once more. Recall that t was already isolated in the previous calculations. This result will be reused.
t = d/s
▼
Substitute values and evaluate
t= 70
Hence, the smartphone will lose connection, once again, after 70 minutes of traveling. Taking into account that it will recover connection after the first 30 minutes, it can be concluded that it will remain connected for 70-30=40 minutes in total.
Solving the Example by Using an Absolute Value Equation
It is known that Davontay's distance from the next village is 150 kilometers. Since the train is moving, his distance is decreasing each minute by 3 kilometers. Using this information, Davontay's distance from the city can be written in terms of the time variable t. distance = 150-3t However, since the distance from the village is a length, it must be a non-negative quantity. This can be be assured by taking the absolute value of the expression. distance = |150-3t| Next, the times when this distance between the train and the city is exactly 60 kilometers should be calculated. This can be done by setting up the following equation. |150-3t| = 60 Since the constant at the right-hand side of the equation is positive, solving the equation is equivalent to solving two individual equations.
Now, the first equation will be solved.
150-3t=- 60
t= 70
Therefore, one of the solutions to the original absolute value equation is t=70. Now, the remaining equation will be solved.
150-3t= 60
t= 30
Consequently, the train will be 60 kilometers away from the city on two occasions. Once, after traveling for 30 minutes while approaching the village. Secondly, after traveling 70 minutes having already passed, and moving away from the village. Therefore, the phone will stay connected for a total of 70-30=40 minutes.
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Example
Using Absolute Value Equations to Define a Range of Prices
Davontay is stoked to buy this video game console that is energized by an eco-friendly battery — everyone wants one. Davontay has $280 saved for this console, but its average selling price is $350. Shopping online, he uses a search program designed to find discounts. He finds that the differences in prices can be modeled with an absolute value equation.
The solutions for this absolute value equation represent the minimum and maximum prices for the console found online. Davontay is planning to buy it at the lowest price. How much more money does Davontay need to save so he can afford the lowest price found online according to this model?
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Hint
The value on the right-hand side of the equation is positive. Therefore, solving the absolute value equation is equivalent to solving two individual equations.
Solution
First, the absolute value equation will be solved to find the minimum selling price. Note that the absolute value expression is already isolated on the left-hand side of the equation, and the quantity on the right-hand side is positive. Therefore, solving this absolute value equation is equivalent to solving two individual one-step equations.
As has been determined, according to the model, the minimum price at which the console can be purchased online is $305. The difference between the minimum price and what Dylan has saved $285 is $25. Therefore, Davontay needs to save $25 more to be able to buy the game console he has long awaited.
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Closure
Using an Absolute Value Equation to Model Battery Performance
A university is developing an eco-friendly battery for tablets called Flora
that uses no harmful chemicals for the environment. After running some tests, the following number line describes the results. Fully charged, the point at 10 indicates the average amount of hours a Flora battery lasts. The points at 8 and 12 indicate the minimum and maximum performance times, respectively.
The university wants to report the specifications of the Flora battery's performance by using an algebraic expression. Let x represent the number of hours the battery can be used when fully charged. Then, find an equation that models the situation and whose solutions are the minimum and the maximum hours the battery can last.
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Hint
What is the distance from the average performance value to the minimum and maximum values? Use this distance to form an absolute value equation.
Solution
To set up an absolute value equation having the required maximum and minimum values as the solutions, it is useful to identify what is the distance from them to the average value on the number line.
As it can be seen from the diagram above, the average value of 10 is two units away from both the minimum and the maximum performance values. Note that the distance of an unknown value x from 10 in a number line can be calculated as a difference of those values. cc if x ≥ 10 & Distance: &x-10 & Example& x = 12 & Distance: &12-10 = 2 if x <10 & Distance: &10-x & Example& x = 8 & Distance: &10-8 = 2 Next, to write the distance in the form of an absolute value expression, the inequalities will be rewritten to have zeros on the right-hand sides resemble the definition for the absolute value of a quantity. x≥ 10 &⇔ x-10≥ 0 [1em] x < 10 &⇔ x-10 < 0 Also, note that 10-x is equivalent to - (x-10). All of these observations can be summarized in the following manner. Distance= x-10 & if x - 10 ≥ 0 - (x-10) & if x - 10 < 0 Notice that this formula for the distance is now very similar to the definition of the absolute value of a number. Therefore, all the previous information imply that the distance for a value x from 10 is |x-10|. Distance=|x -10| Finally, since it is known that the distance to the desired values is 2, the required equation can be set up. 2=|x-10| ⇕ |x-10|=2 Now that the Flora battery has been tested and the performance results can be reported using a formal mathematical expression, the university can let everyone know about Flora to power devices in a cleaner way!
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Solving Absolute Value Equations
Exercise 2.1
Izabella is trying to convince her friend Kriz that |a| + |b| and |a + b| have the same output since everything becomes positive at the end. Kriz says that this cannot be right. Who is correct?
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We want to verify if what Izabella says is true. To do so, we will examine different cases for the values of a and b, when one both are positive, when both are negative, and when the numbers have opposite signs.
Both Numbers Are Positive
First, we will examine the case where both values are positive. In this case, we will substitute two arbitrary positive numbers for a and b. Let the numbers be a= 5 and b= 3.
Now we will substitute these values into the other expression.
We can see that Izabella is right when the two numbers are positive.
Both Numbers Are Negative
This time we will examine the case where both values are negative. Similarly to the previous case, we will substitute two arbitrary negative numbers for a and b. This time, the numbers will be a= -5 and b= -3.
And now we will substitute these values into the other expression.
We can also see that Izabella is right when the two numbers are negative.
Numbers Have Opposite Signs
Finally, let's try the expressions when one value is positive and the other is negative. We will use the arbitrary values a=5 and b=-3.
Now we will calculate the value of |a + b|.
Since we found example values of a and b that make the expressions different, the expressions cannot be equal to each other. This means that Izabella's claim is not always true.
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Solution
The equation below shows the graphed equation. |x-$ $|= $ $
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{"type":"choice","form":{"alts":["a","b","c","d"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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An absolute value equation can be written following a format. |x- l|= r In this equation, l is the midpoint and r is the distance from the values of x to the midpoint. From the diagram, we can see that the midpoint, the point that divides the distance between the solutions into two equal parts, is located at b.
Now we know that the value of l from the equation is equal to b. Therefore, begin to fill in the blanks by writing b inside the red box. |x-$b$|= $ $
Let's look at the absolute value equation again. |x- l|= r The value on the right-hand side of the absolute value equation r gives the distance from the midpoint to the equation's solutions. Let's consider our diagram again.
We can see that the distance between the midpoint and either solution is d. This means that the value of r is d. Therefore, to complete the exercise, we should write d in the purple box. |x-$b$|= $d$
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Solution
|
If x^2=a^2, then |x| is equal to |a|. |
{"type":"choice","form":{"alts":["always","sometimes","never"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
|
If x and y are real numbers, then |x-y| is equal to |y-x|. |
{"type":"choice","form":{"alts":["always","sometimes","never"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
|
For any real number r, the equation |x-5|=r will have two solutions. |
{"type":"choice","form":{"alts":["always","sometimes","never"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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Recall that a square root of a number raised to the second power is equal to the absolute value of that number. sqrt(a^2)=|a| Let's use this when we calculate the square root of both sides of the equation.
Since we have an equals sign between |x| and |a|, we know that we should fill in the blank with always.
If x^2=a^2, then |x| is always equal to |a|.
The absolute values of any real number and its opposite are always the same. This is because they are equidistant from 0 on a number line.
| a| = |- a|
This must also mean that a-b is equal to the absolute value of its opposite expression, which we can write as -(a-b). Let's simplify this expression first.
We found that b-a is the opposite expression of a-b. We can now write the following absolute value equation. |a-b|=|b-a| Since we have an equals sign between |a-b| and |b-a|, we can fill in the blank in the sentence with the word always.
If x and y are real numbers, then |x-y| is always equal to |y-x|.
We are given the following absolute value equation.
|x-5|=r
The number of solutions depends on the sign of r.
- If r is negative, there are no solutions. This is because the absolute value of any number is always a non-negative number.
- If r is 0, there will be one solution.
- If r is positive, there will be two solutions because x-5 can equal either r or - r.
Therefore, we can fill in the blank in the statement with the word sometimes.
For any real number r, the equation |x-5|=r will sometimes have two solutions.
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Solution
While standing at the bus stop, Magdalena sees a car approaching. When she sees it, the car is 200 feet away from her and is approaching at a speed of 40 feet per second. For how many seconds is the car within 80 feet of Magdalena?
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We need to figure out for how long was the car within 80 feet of Magdalena. To do so, we need to find the time at which the car was 80 feet behind Magdalena and the time at which the car was 80 feet in front of her. The difference between these time gives us the total amount of time in seconds that the car is within 80 feet of Magdalena. We can illustrate this with a diagram.
We can write an absolute value equation to model when the car is both 80 feet in front and 80 feet behind Magdalena. Such an equation generally follows the following format. |x- midpoint|= distance Let's define the distance and midpoint.
Distance
The right-hand side of the equation is the distance between the equation's midpoint and its solutions. The midpoint divides the distance into two equal segments. From the diagram, we can see that the right-hand side of this equation should be 80 feet. Let's add this value to the equation. |x- midpoint|= 80
Midpoint
Let's adjust our diagram to place it on a coordinate plane. The car starts out 200 feet away from Magdalena. Because distance is always non-negative, if we place Magdalena at the midpoint of the equation, we should place her at least the same distance from the y-axis. This means that her coordinates are ( 200,0), and the car's initial position is at (400,0).
Now we will substitute 200 for the midpoint in the absolute value equation. |x- 200|= 80
Replacing x
The distance from the midpoint is given by |x-200|, where x represents the number of feet the car has traveled after t seconds. The distance the car travels can be calculated by multiplying the time the car has traveled t by the speed of the car, which was given as 40 feet per second. x = 40t Let's substitute 40t for x and complete our absolute value equation. |40t-200|=80
Solving the Equation
Let's solve the absolute value equation.
We found that the car is 80 feet behind Magdalena 3 seconds after she started measuring the time, and that the car was 80 feet in front of Magdalena 7 seconds after she started counting. The difference between these times indicates the total amount of time that the car is within 80 feet of her. This means that the car is within 80 feet of her for 7-3=4 seconds.
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Solving Absolute Value Equations
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