Unlock the Secrets of Solving Absolute Value Equations: Different Methods and Properties of Equality

Property	Algebraic Representation
Non-negativity	$∣ a ∣ \geq 0$
Symmetry	$∣ - a ∣ = ∣ a ∣$
Idempotence	$∣ ∣ a ∣ ∣ = ∣ a ∣$
Positive-definiteness	$∣ a ∣ = 0 \Leftrightarrow a = 0$
Identity of Indiscernibles	$∣ a - b ∣ = 0 \Leftrightarrow a = b$
Multiplicativity	$∣ a b ∣ = ∣ a ∣ \cdot ∣ b ∣$
Preservation of Division	$∣ ∣ ∣ ∣ \frac{a}{b} ∣ ∣ ∣ ∣ = \frac{∣ a ∣}{∣ b ∣} if b \neq = 0$
Subadditivity	$∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$
Triangle Inequality	$∣ a - b ∣ \leq ∣ a - c ∣ + ∣ c - b ∣$

Property

Algebraic Representation

Non-negativity

∣ a ∣ \geq 0

Symmetry

∣ - a ∣ = ∣ a ∣

Idempotence

∣ ∣ a ∣ ∣ = ∣ a ∣

Positive-definiteness

∣ a ∣ = 0 \Leftrightarrow a = 0

Identity of Indiscernibles

∣ a - b ∣ = 0 \Leftrightarrow a = b

Multiplicativity

∣ a b ∣ = ∣ a ∣ \cdot ∣ b ∣

Preservation of Division

∣ ∣ ∣ ∣ \frac{a}{b} ∣ ∣ ∣ ∣ = \frac{∣ a ∣}{∣ b ∣} if b \neq = 0

Subadditivity

∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣

Triangle Inequality

∣ a - b ∣ \leq ∣ a - c ∣ + ∣ c - b ∣

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	$- 2, 2, - 6, 6$

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

- 2, 2, - 6, 6

$2 x + 4 = 16$	$2 x + 4 = - 16$
$2 x = 12$	$2 x = - 20$
$x = \frac{1 2}{2}$	$x = \frac{- 2 0}{2}$
$x = 6$	$x = - 10$

2 x + 4 = 16

2 x + 4 = - 16

2 x = 12

2 x = - 20

x = \frac{1 2}{2}

x = \frac{- 2 0}{2}

x = 6

x = - 10

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?

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.

Consider the following graph.

a number line showing an absolute value equation

The equation below shows the graphed equation.

∣ ∣ ∣ ∣ x - ? ∣ ∣ ∣ ∣ = ?

What variable should we write in the red box?

What variable should we write in the purple box?

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 $x^{2} = a^{2},$ then $∣ x ∣$ is $\underline{always}$ equal to $∣ a ∣ .$

If $x$ and $y$ are real numbers, then $∣ x - y ∣$ is $\underline{always}$ equal to $∣ y - x ∣ .$

For any real number $r$ , the equation $∣ x - 5 ∣ = r$ will $\underline{always}$ 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.

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.

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?

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.

Solving Absolute Value Equations

Catch-Up and Review

Modeling the Performance of a Battery

Why and How to Define an Always Positive Quantity

Absolute Value

Absolute Value Properties

Simplifying Absolute Value Expressions

Absolute Value Equations

Number of Solutions

Solving Absolute Value Equations Algebraically

Extra

Practice Solving Absolute Value Equations

Using Absolute Value Equations to Model Real Life Situations

Hint

Solution

Solving the Example by Using a Number Line

Solving the Example by Using an Absolute Value Equation

Using Absolute Value Equations to Define a Range of Prices

Hint

Solution

Using an Absolute Value Equation to Model Battery Performance

Hint

Solution

Both Numbers Are Positive

Both Numbers Are Negative

Numbers Have Opposite Signs

Distance

Midpoint

Replacing x

Solving the Equation

Solving Absolute Value Equations

Recommended exercises

	10 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions