Lesson Objectives

Dear learner;

At the end of the lesson you will be able to:

• Solve absolute value inequalities.
• Graph inequalities involving absolute value.
• determine the solution sets for absolute value inequalities.
• use the rules for solving absolute value inequalities.

Key Terms

• Absolute value inequality
• solution set

Brainstorming Question

1. Marta is planning to buy vegetables at the local market in Addis Ababa. She wants to spend no more than 50 birr on vegetables, but she is willing to tolerate a small fluctuation in the total amount due to varying prices. She estimates that the actual amount she spends will be within 5 birr of her budgeted amount. Write an inequality to represent the total amount Marta can spend on vegetables and solve it

Answer:
The inequality representing Marta’s spending can be written as:
$[ |x – 50| \leq 5 ]$
where ( x ) is the total amount she spends.

To solve:

1. Break it into two inequalities:
   $[ -5 \leq x – 50 \leq 5 ]$
2. Add 50 to each part:
   $[ -5 + 50 \leq x \leq 5 + 50 ]$
   $[ 45 \leq x \leq 55 ]$

Marta can spend between 45 birr and 55 birr on vegetables.

2. Two towns, Debre Markos and Bahir Dar, are located in Ethiopia. The distance between them is approximately 200 kilometers, but due to varying routes, the actual distance can fluctuate by up to 15 kilometers. Write an inequality to represent the possible distances between the two towns and solve it.

Answer:
The inequality representing the possible distances can be written as:
$[ |d – 200| \leq 15 ]$
where ( d ) is the actual distance between the towns.

To solve:

1. Break it into two inequalities:
   $[ -15 \leq d – 200 \leq 15 ]$
2. Add 200 to each part:
   $[ -15 + 200 \leq d \leq 15 + 200 ]$
   $[ 185 \leq d \leq 215 ]$

The distance between Debre Markos and Bahir Dar can vary between 185 kilometers and 215 kilometers.

3.1. Absolute Value inequality

• The absolute value of a number is its distance from zero on the number line, always resulting in a non-negative value.
• The distance of a number ( x ) from zero on the number line, always resulting in a non-negative value. For example, ( |3| = 3 ) and ( |-3| = 3 )

When dealing with absolute value inequalities, there are two main types of expressions to consider:

• $|x| < a$ or $|x| ≤ a$: This type of inequality states that the absolute value of x is less than (or less than or equal to) a. In this case, x can take on any value within the interval $(-a, a)$ for $|x| < a$ or $[-a, a]$ for $|x| ≤ a$.

Graphically: $(-a, a)$ for $|x| < a$

Graphically: $[-a, a]$ for $|x| ≤ a$

• $|x| > a$ or $|x| ≥ a$: This inequality indicates that the absolute value of x is greater than (or greater than or equal to) a. The solution set for this type of inequality consists of

a. $|x| > a$:
– This represents all values of x such that $x > a$ or $x < – a$):
– On the number line, this would be represented by two open intervals: $(negative infinity, -a)$ and $(a, infinity)$.

b. $|x| ≥ a$:
– This represents all values of x that is $(x \ge a or x \le – a)$:.
– On the number line, this would be represented by two closed intervals: $(negative infinity, -a]$ and $[a, infinity)$.

Graphically

Example 1:

Graph the following absolute value inequality on a number line.
a. $|x| ≤ 1$ b. $|x| > 1$

Solution:
a. $|x| ≤ 1$ are the numbers that satisfy $−1 < x < 1$.

b. $|x| > 1$ are the numbers that satisfy $x < −1$ or $x > 1$.

.When expressing inequalities involving absolute value, the following rules apply:

1. For $|k| < 𝑐$:
The solutions are the numbers that satisfy $- c < k < c$.

2. For $|k| > c$:
The solutions are the numbers that satisfy $k > c$ or $k < – c$.

These rules remain valid if < is replaced by ≤ and > is replaced by ≥. By applying these rules, you can effectively solve inequalities involving absolute value and determine the appropriate range of solutions.

3.2. Inequalities Involving Absolute Value

Inequalities where the variable is within absolute value brackets, such as (|x – a| < b) or (|x – a| > b). These express distances and are solved by splitting into separate linear inequalities.

Example 1: Find the solutions to the given absolute value inequalities.

a) $|x – 2| < 7$

b) $|6 − 3x| ≤ 0$

c) $|x + 12| > 4$

d) $|2x − 3| ≥ −3$

Solution

a) Rewrite the absolute value inequality without the absolute value. This gives us two separate inequalities: – 7 < x – 2 < 7.

1. To isolate x, add 2 to each part of the inequality:$− 7 + 2 < x – 2 + 2 < 7 + 2$
2. Simplify each part:$ − 5 < x < 9$

b) $|6 – 3x| ≤ 0$. Thus the solution can be calculated as $0 ≤ 6 – 3x ≤ 0$, subtract 6 to each term. It will give as $- 6 ≤ – 3x ≤ – 6$, and finally divide each term by – 3. therefore, the solution for $2 ≤ x ≤ 2$ will be $x = 2$.

or

To solve the inequality $∣6 – 3x∣ ≤ 0$, we need to analyze the properties of absolute values and inequalities. Here is the step-by-step process:

1. Understand the absolute value inequality:$∣6 – 3x∣ ≤ 0$
2. Recall that the absolute value of any expression is always non-negative, meaning $∣a∣ ≥ 0$ for any a. The only way $∣a∣ ≤ 0$ is if $a = 0$.
3. Therefore, to satisfy ∣6 – 3x∣≤0, the expression inside the absolute value must be zero: $6 – 3x = 0$.
4. Solve the equation $6 – 3x = 0$.
5. Subtract 6 from both sides to isolate the term with x:$ − 3x = −6$.
6. Divide both sides by -3 to solve for $x: x = 2$

Therefore, the solution to the inequality $∣6 – 3x∣ ≤ 0$ is :$x = 2$.

In this case, since $∣6 − 3x|$ is zero only when $x = 2$, the solution is a single value $x = 2$.

c) To solve the inequality $∣x+12∣ > 4$, we need to break it down into two separate cases because the absolute value represents a distance that can be either positive or negative. Here’s the step-by-step process:

1. Understand the absolute value inequality:$∣x+12∣>4$
2. The absolute value inequality $∣a∣ > b$ (where b >0) can be rewritten as two separate inequalities: $a > b \quad \text{or} \quad a < -b$
3. Apply this to the given inequality $|x + 12| > 4: x + 12 > 4 \quad \text{or} \quad x + 12 < -4$
4. Solve each inequality separately:
   • For $x + 12 > 4$: Subtract 12 from both sides: $x> 4 – 12$ Simplify: $x > -8$
   • For $x + 12 < -4$: Subtract 12 from both sides: $x< − 4 – 12$ Simplify: $x < -16$
5. Combine the two separate solutions:$x > -8 \quad \text{or} \quad x < -16$

Therefore, the solution to the inequality $∣x + 12∣ > 4$ is: $x > -8 \quad \text{or} \quad x < -16$

In interval notation, this is expressed as:$(-\infty, -16) \cup (-8, \infty)$

d) To solve the inequality $|2x – 3| \geq -3$, let’s start by analyzing the properties of absolute values and the inequality.

1. Understand the absolute value properties:∣a∣ ≥ 0 for any real number a, $|a| \geq 0 \quad \text{for any real number }$. This means the absolute value of any expression is always non-negative. Therefore:$|2x – 3| \geq 0$
2. Compare with the given inequality: The given inequality is:$|2x – 3| \geq -3$
3. Analyze the given inequality: Since $|2x – 3|$ is always non-negative and non-negative values are always greater than or equal to any negative number, the inequality $|2x – 3| \geq -3$ is always true for any real number x.
4. Conclusion: The inequality $|2x – 3| \geq -3$ holds true for all real numbers because the absolute value of any expression is always non-negative and thus always greater than or equal to -3.

Therefore, the solution to the inequality $|2x – 3| \geq -3$ is: $x \in \mathbb{R}$

In other words, the inequality is true for all real numbers.

Example 2:

In a café in Addis Ababa, the barista wants to keep the temperature of the coffee between 70°C and 80°C. However, the coffee machine has a margin of error of ±3°C. Express the acceptable temperature range for the coffee machine using an absolute value inequality and find the range of temperatures the machine should maintain.

Solution
The acceptable temperature range can be expressed as:
$[ |T – 75| \leq 5 ]$
where ( T ) is the temperature of the coffee.

To solve:

1. Break it into two inequalities:
   $[ -5 \leq T – 75 \leq 5 ]$
2. Add 75 to each part:
   $[ -5 + 75 \leq T \leq 5 + 75 ]$
   $[ 70 \leq T \leq 80 ]$

The machine should maintain a temperature between 70°C and 80°C

Here are the key terms for a lesson on inequalities involving absolute values: