Lesson Objectives

Dear learner;

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

• Express the interval notation in inequalities and on a number line.
• Add, subtract, multiply and divide by a constant from both sides of an inequality to solve linear inequalities.
• Multiply or divide both sides of an inequality by a positive constant, and verify that the direction of the inequality remains the same.
• Adjust an inequality by multiplying or dividing both sides by a negative constant, and confirm that the direction of the inequality reverses.
• Determine the solutions of linear inequalities using interval notation and on a number line to clearly represent the set of possible values.

Key Terms

• linear inequality
• open interval
• closed interval
• half open interval

Brainstorming Questions

1. Maria is planning her monthly budget. She has set aside 500 Birrfor entertainment expenses and wants to divide it between movies, dining out, and other activities. She finds that if she spends more than 150 Birr on movies, she must reduce her dining out budget to ensure her total entertainment expenses stay within her budget.

2. If Maria spends 180 Birr on movies, what is the maximum amount she can spend on dining out if her total budget is 500 Birr? Use inequalities to determine the answer.

Answer:

1. Determine the inequality for total expenses: Let (m) be the amount spent on movies, and (d) be the amount spent on dining out. Given that Maria’s total budget is 500 Birrr, and if she spends more than 150 Birr on movies, the inequality representing the total budget constraint is:
   $m + d \leq 500$
2. Substitute ( m = 180 ) into the inequality:
   $180 + d \leq 500$
3. Solve for ( d ):
   $d \leq 500 – 180$
   $d \leq 320$
   Conclusion: If Maria spends 180Birrr on movies, she can spend up to 320 Birrr on dining out to stay within her 500 Birr budget.

1.1. Revision on Linear Inequalities in One Variable

Linear Inequalities in One Variable: Inequalities of the form $(ax + b < c)$, $(ax + b \leq c)$, $(ax + b > c)$, or $(ax + b \geq c)$, where (x) is the variable and (a), (b), and (c) are constants. The solution is the range of (x) values that satisfy the inequality.

Dear learner; Some Properties of Inequalities

If $( a > b )$, then $( a + c > b + c )$
If $( a > b )$, then $( a – c > b – c )$
If $( a > b )$ and $( m > 0 )$, then $( ma > mb )$ and $( \frac{a}{m} > \frac{b}{m} )$
If $( a > b )$ and $( m < 0 )$, then $( ma < mb )$ and $(\frac{a}{m} < \frac{b}{m} )$

Example 1: Solve the following inequalities. a) $\frac{1}{3}x \geqslant 4$ b) $2 x – 3 ≥ 1$ c) $5𝑥 − 3(𝑥 − 1) < 5$

Solution:

a) $\frac{1}{3} x \geqslant 4$ (multiply both sides by 3)

$x \geqslant 12$

b) $2 x – 3 ≥ 1$ (add three to both sides)

$2 x ≥ 4$ (divide both sides by 2)

$ x ≥ 2$

c) $5x – 3(x – 1) < 5$

$5x – 3x + 3 < 5$

$2x + 3 < 5$

$2x < 2$

$x < 1$

Note

• The symbol $[𝑎, 𝑏]$ represents a closed interval with endpoints 𝑎 and 𝑏. This notation denotes the set of all real numbers 𝑥 such that $a \leqslant x \leqslant b$ (𝑎 is less than or equal to 𝑥, and 𝑥 is less than or equal to 𝑏).
• The symbol $(𝑎, 𝑏)$ signifies an open interval with endpoints 𝑎 and 𝑏. This notation indicates the set of all real numbers 𝑥 such that $a < x < b$ (𝑎 is less than 𝑥, and 𝑥 is less than 𝑏).
• The symbols $[𝑎, 𝑏)$ and $(𝑎, 𝑏]$ represent half-open intervals. This notation indicates the set of all real numbers 𝑥 such that $a \leqslant x < b$ (𝑎 is less than or equal to 𝑥, and 𝑥 is less than 𝑏).
• The symbols $(𝑎, 𝑏]$ represent half-open intervals. This notation indicates the set of all real numbers 𝑥 such that $a < x \leqslant b$ (𝑎 is less than 𝑥, and 𝑥 is less than or equal to 𝑏).
• These interval notations are essential in mathematics for defining ranges of values and understanding the relationships between different sets of real numbers.

Example 2: Express the following using the notation of intervals.

a)

Answer:- a. [1, 6]

b)

Answer: b. [−3, 2)

Example 3: Express the following intervals on a number line.
a. $(−1,3)$ b. $[2, 5)$

Solution: a) (−1,3)

b. $[2, 5)$

Example 4:

A company ships products and charges for shipping based on weight. For weights greater than 10 kg but less than or equal to 20 kg, the company charges 5 Birr per kg. If the weight of a package is ( w ) kg, what is the maximum weight that still incurs a shipping cost of 100 US Dollar or less? What is the maximum weight ( w ) that will result in a total shipping cost of 100 Birr or less?

Solution:

Set up the inequality for the shipping cost: The cost per kg is 5 Birrr . If the total cost must be 100 Birrr or less:
$5w \leq 100$

Solve for ( w ):
$w \leq \frac{100}{5}$
$w \leq 20$
Conclusion: The maximum weight (w) that will result in a shipping cost of 100 Birrr or less is 20 kg.

2. A laboratory needs to maintain a temperature range for a chemical experiment. The temperature ( T ) must be higher than 25°C but not exceed 35°C for the reaction to occur correctly. If the temperature control system allows a maximum temperature adjustment of ±5°C from its current setting of 30°C, what are the limits for the adjusted temperature to ensure the reaction conditions are met?

Solution:

Set up the inequality for the temperature adjustment:

The current temperature is 30°C, and the allowable adjustment is ±5°C, so the adjusted temperature ( T ) can be:
$30 – 5 \leq T \leq 30 + 5$
$25 \leq T \leq 35$

Verify the range against the required conditions:

The required temperature range for the reaction is $25 < T < 35$. Since the adjusted range exactly fits this requirement:
$25 \leq T \leq 35$
Conclusion: The adjusted temperature must be between 25°C and 35°C to meet the reaction conditions