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Lesson 3-4: Inequalities & Applications

Scenario: Budget & Grade Planning - Learn to solve inequalities and represent solutions on number lines!

Duration: 60-75 minutesScenario: Budget & Grade Planning

Learning Objectives

  • Solve inequalities using inverse operations
  • Represent inequality solutions on number lines
  • Apply inequalities to budget and grade planning
  • Understand the difference between < and ≤ symbols

Grade Requirement Problem

The Problem

Sarah has taken 3 tests with scores of 82, 78, and 85. She needs an average of at least 85 to get an A in the class. What score does she need on the fourth test?

Setting up the inequality:
Average = (82 + 78 + 85 + x) ÷ 4
(82 + 78 + 85 + x) ÷ 4 ≥ 85

Solving the Inequality

Step 1: Simplify the numerator

(245 + x) ÷ 4 ≥ 85

Step 2: Multiply both sides by 4

245 + x ≥ 340

Step 3: Subtract 245 from both sides

x ≥ 95

Answer: Sarah needs at least 95 on the fourth test.

Number Line Representation

Representing x ≥ 95

Number Line:

90
92
94
96
98

The filled circle at 95 means "include 95", and the arrow shows "all numbers greater than or equal to 95".

Inequality Symbols

Strict Inequalities

< (less than)
> (greater than)
Use open circles on number line

Non-strict Inequalities

≤ (less than or equal to)
≥ (greater than or equal to)
Use filled circles on number line

Budget Problem

The Problem

Tom has a budget of $200 for school supplies. He has already spent $85. He wants to buy notebooks that cost $3 each. How many notebooks can he afford?

Setting up the inequality:
Amount spent + (Cost per notebook × Number of notebooks) ≤ Budget
85 + 3x ≤ 200

Solving the Inequality

Step 1: Subtract 85 from both sides

3x ≤ 115

Step 2: Divide both sides by 3

x ≤ 38.33...

Answer: Tom can afford at most 38 notebooks (since he can't buy a fraction of a notebook).

Practice Problems

Problem 1: Simple Inequality

Solve: 2x - 5 < 11

Your solution:

Problem 2: Grade Problem

A student has scores of 88, 92, and 85. What score is needed on the fourth test to have an average of at least 90?

Your calculation: