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| 10 Theory slides |
| 10 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.
Here are a few practice exercises before getting started with this lesson.
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.
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.
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 | ∣∣∣∣ba∣∣∣∣=∣b∣∣a∣ if b=0 |
Subadditivity | ∣a+b∣≤∣a∣+∣b∣ |
Triangle Inequality | ∣a−b∣≤∣a−c∣+∣c−b∣ |
Practice simplifying absolute value expressions by using the following applet.
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.
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 |
∣∣∣x2−4∣∣∣=2 | Four | -2,2,-6,6 |
Solve each linear equation found in Step 2 separately.
2x+4=16 | 2x+4=-16 |
---|---|
2x=12 | 2x=-20 |
x=212 | x=2-20 |
x=6 | x=-10 |
After writing the absolute value equation in the form ∣ax+b∣=c, where a, b, and c are constants, check the value of c.
Practice solving absolute value equations by using the following applet. Indicate which number line represents the solution set of the given equation.
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?
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.
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.
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 value on the right-hand side of the equation is positive. Therefore, solving the absolute value equation is equivalent to solving two individual equations.
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.
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.
What is the distance from the average performance value to the minimum and maximum values? Use this distance to form an absolute value equation.
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.
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.
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.
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.
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.
Consider the following graph.
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-$b$|= $d$
Complete the following statements with always, sometimes or never.
If x2=a2, then ∣x∣ is always equal to ∣a∣. |
If x and y are real numbers, then ∣x−y∣ is always equal to ∣y−x∣. |
For any real number r, the equation ∣x−5∣=r will always have two solutions. |
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.
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.
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.
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
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
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
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.