Solved Exercises on Linear Inequality Word Problems in Integrated Math 1

Master linear inequality word problems: systems, graphs, real-world applications, and solution methods through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Budget constraint problem
Exercise 1
Sarah has $50 to spend on books and magazines. Books cost $8 each and magazines cost $3 each. Write an inequality representing the possible combinations of books (x) and magazines (y) she can buy.
Definition:

Linear inequality: An inequality that represents a relationship between variables where the highest degree is 1.

Word problem translation: Convert verbal constraints into mathematical inequalities.

Translation method:
  1. Identify variables (x, y)
  2. Identify costs per item
  3. Identify total budget constraint
  4. Set up inequality using ≤, ≥, <, or >
Variables
x = books, y = magazines
Costs
$8x + $3y
Budget
≤ $50
Step 1: Define variables

Let x = number of books, y = number of magazines

Step 2: Calculate total cost

Total cost = (cost per book × number of books) + (cost per magazine × number of magazines)

Total cost = 8x + 3y

Step 3: Set up budget constraint

Since Sarah cannot spend more than $50: 8x + 3y ≤ 50

Step 4: Interpret the inequality

All points (x,y) that satisfy 8x + 3y ≤ 50 represent possible purchases

\(8x + 3y \leq 50\)
Final answer:

The inequality representing Sarah's purchase options is: 8x + 3y ≤ 50

This means she can buy x books and y magazines as long as the total cost doesn't exceed $50.

Applied rules:

• Budget constraint: Total spending ≤ available money

• Linear combination: Sum of individual costs forms the inequality

• Non-negative values: x ≥ 0 and y ≥ 0 (can't buy negative items)

2 Distance and time problem
Exercise 2
A delivery truck can carry at most 2000 pounds. Packages weigh 25 pounds each and boxes weigh 40 pounds each. Write an inequality showing the possible combinations of packages (p) and boxes (b) that can be loaded.
Definition:

Capacity constraint: An upper limit on the total weight, volume, or quantity that can be handled.

Weight inequality: Total weight ≤ maximum capacity

Variables
p = packages, b = boxes
Weights
25p + 40b
Capacity
≤ 2000 lbs
Step 1: Define variables

Let p = number of packages, b = number of boxes

Step 2: Calculate total weight

Total weight = (weight per package × number of packages) + (weight per box × number of boxes)

Total weight = 25p + 40b

Step 3: Set up capacity constraint

Since the truck can carry at most 2000 pounds: 25p + 40b ≤ 2000

Step 4: Simplify if possible

We can divide by 5: 5p + 8b ≤ 400

\(25p + 40b \leq 2000\)
Final answer:

The inequality representing the truck's load capacity is: 25p + 40b ≤ 2000

Or simplified: 5p + 8b ≤ 400

Applied rules:

• Capacity constraint: Total weight ≤ maximum capacity

• Linear inequality: Forms a boundary in the coordinate plane

• Non-negative constraints: p ≥ 0 and b ≥ 0

3 Nutrition constraint problem
Exercise 3
A diet plan requires at least 1200 calories per day. An apple provides 95 calories and a granola bar provides 150 calories. Write an inequality for the possible daily combinations of apples (a) and granola bars (g) to meet the calorie requirement.
Definition:

Minimum requirement: An inequality stating that the total must be greater than or equal to a specified value.

Nutrition constraint: Total nutrition ≥ minimum requirement

Variables
a = apples, g = granola bars
Calories
95a + 150g
Requirement
≥ 1200 cal
Step 1: Define variables

Let a = number of apples, g = number of granola bars

Step 2: Calculate total calories

Total calories = (calories per apple × number of apples) + (calories per granola bar × number of granola bars)

Total calories = 95a + 150g

Step 3: Set up minimum requirement

Since the diet requires at least 1200 calories: 95a + 150g ≥ 1200

Step 4: Simplify if possible

We can divide by 5: 19a + 30g ≥ 240

\(95a + 150g \geq 1200\)
Final answer:

The inequality representing the calorie requirement is: 95a + 150g ≥ 1200

Or simplified: 19a + 30g ≥ 240

Applied rules:

• Minimum requirement: Total ≥ required minimum

• Greater-than-or-equal: Used for "at least" scenarios

• Non-negative constraints: a ≥ 0 and g ≥ 0

Rules and methods, laws,...
\(ax + by \leq c \text{ or } ax + by \geq c\)
Linear Inequality Standard Form
Key definitions:

Linear inequality: An inequality containing variables of degree 1

Constraint: A condition limiting possible values

Feasible region: Area satisfying all constraints

Complete methodology:
  1. Read the problem carefully: Identify constraints and objectives
  2. Define variables: Assign letters to unknown quantities
  3. Translate words to math: Convert verbal constraints to inequalities
  4. Identify inequality signs: Use ≤ for "at most", ≥ for "at least"
  5. Solve or graph: Find feasible region
Tip 1: "At most" means ≤, "at least" means ≥, "less than" means <, "more than" means >
Tip 2: Always consider non-negative constraints: x ≥ 0, y ≥ 0
Tip 3: Check your solution by substituting values back into the original problem
Tip 4: Draw a graph to visualize the feasible region
Common errors: Using wrong inequality signs, forgetting non-negative constraints, misinterpreting "at most" vs "at least"
Exam preparation: Practice translating words to inequalities, work on graphing systems, understand constraint interpretation
Formulas to know by heart:

• Budget constraint: Total cost ≤ available money

• Capacity constraint: Total weight ≤ maximum capacity

• Minimum requirement: Total ≥ required minimum

• Standard form: ax + by ≤ c or ax + by ≥ c

Solution: Exercises 4 to 5
4 Attendance constraint problem
Exercise 4
A school bus can accommodate at most 45 students. Students from two classes want to take the trip: Class A has x students and Class B has y students. Write an inequality representing the possible combinations of students that can go on the trip.
Definition:

Capacity constraint: An upper limit on the number of people or items that can be accommodated.

Attendance inequality: Total attendance ≤ maximum capacity

Variables
x = Class A students, y = Class B students
Total
x + y
Capacity
≤ 45
Step 1: Define variables

Let x = number of students from Class A, y = number of students from Class B

Step 2: Calculate total attendance

Total students = number from Class A + number from Class B

Total students = x + y

Step 3: Set up capacity constraint

Since the bus can accommodate at most 45 students: x + y ≤ 45

Step 4: Consider additional constraints

We also need x ≥ 0 and y ≥ 0 (can't have negative students)

\(x + y \leq 45\)
Final answer:

The inequality representing the bus capacity is: x + y ≤ 45

With additional constraints: x ≥ 0 and y ≥ 0

Applied rules:

• Capacity constraint: Total ≤ maximum allowed

• Linear combination: Simple addition of variables

• Non-negative constraints: Variables must be ≥ 0

5 Time constraint problem
Exercise 5
A student has 2 hours (120 minutes) to study for two subjects: Math and Science. She wants to spend at least 30 minutes on Math and at least 20 minutes on Science. Let x be minutes spent on Math and y be minutes spent on Science. Write a system of inequalities describing her study plan.
Definition:

System of inequalities: Multiple inequalities that must all be satisfied simultaneously.

Time constraint: Total time ≤ available time

Variables
x = Math minutes, y = Science minutes
Total time
x + y ≤ 120
Min requirements
x ≥ 30, y ≥ 20
Step 1: Define variables

Let x = minutes spent on Math, y = minutes spent on Science

Step 2: Set up total time constraint

Since total study time cannot exceed 120 minutes: x + y ≤ 120

Step 3: Set up minimum requirements

At least 30 minutes on Math: x ≥ 30

At least 20 minutes on Science: y ≥ 20

Step 4: Combine all constraints

System of inequalities: x + y ≤ 120, x ≥ 30, y ≥ 20

\begin{cases} x + y \leq 120 \\ x \geq 30 \\ y \geq 20 \end{cases}
Final answer:

The system of inequalities is: x + y ≤ 120, x ≥ 30, y ≥ 20

This represents all possible study time allocations that meet the requirements.

Applied rules:

• System formation: Multiple constraints form a system

• Time management: Total usage ≤ available time

• Minimum requirements: Individual constraints using ≥

Linear Inequality Word Problems Guide
\(ax + by \leq c \text{ or } ax + by \geq c\)
Linear Inequality Standard Form
Key definitions:

Linear inequality: An inequality containing variables of degree 1

Constraint: A condition limiting possible values

Feasible region: Area satisfying all constraints

Objective function: Function to maximize or minimize within constraints

Complete methodology:
  1. Read the problem carefully: Identify constraints and objectives
  2. Define variables: Assign letters to unknown quantities
  3. Translate words to math: Convert verbal constraints to inequalities
  4. Identify inequality signs: Use ≤ for "at most", ≥ for "at least"
  5. Solve or graph: Find feasible region
Tip 1: "At most" means ≤, "at least" means ≥, "less than" means <, "more than" means >
Tip 2: Always consider non-negative constraints: x ≥ 0, y ≥ 0
Tip 3: Check your solution by substituting values back into the original problem
Tip 4: Draw a graph to visualize the feasible region
Tip 5: Look for key phrases like "cannot exceed", "must be at least", "at most", "minimum"
Common errors: Using wrong inequality signs, forgetting non-negative constraints, misinterpreting "at most" vs "at least"
Exam preparation: Practice translating words to inequalities, work on graphing systems, understand constraint interpretation
Formulas to know by heart:

• Budget constraint: Total cost ≤ available money

• Capacity constraint: Total weight ≤ maximum capacity

• Minimum requirement: Total ≥ required minimum

• Standard form: ax + by ≤ c or ax + by ≥ c

Linear Inequality Systems Visualization
System Analysis: Study Time Optimization
Consider the system of inequalities from Exercise 5:
\(x + y \leq 120\) (total time constraint)
\(x \geq 30\) (minimum Math time)
\(y \geq 20\) (minimum Science time)
This system defines a feasible region for study time allocation.

Analysis: The feasible region shows all possible combinations of study time that satisfy all constraints.

  • Corner points of the feasible region often represent optimal solutions
  • Constraints form boundaries of the feasible region
  • Any point inside or on the boundary satisfies all constraints

Questions & Answers

Question: I'm confused about when to use ≤ versus < in word problems. How do I know which one to choose?

Answer: Great question! The key is in the wording:

  • Use ≤ (less than or equal) when the problem says:
    • "at most" - e.g., "at most $50" means 50 is allowed
    • "cannot exceed" - e.g., "cannot exceed 45 students" means 45 is the maximum allowed
    • "maximum of" - e.g., "maximum of 2000 pounds" means 2000 is acceptable
  • Use < (strictly less than) when the problem says:
    • "less than" - e.g., "less than 50 students" means 50 is NOT allowed
    • "under" - e.g., "under $100" means $100 is not included

In most practical problems, "at most" is more common because limits are typically inclusive of the boundary value.

Question: Why do we need to consider x ≥ 0 and y ≥ 0 in word problems? Can't we just use the main inequality?

Answer: The constraints x ≥ 0 and y ≥ 0 are called "non-negativity constraints" and they're crucial for real-world problems:

  • In our examples, you can't buy negative books, carry negative weight, eat negative food, or study negative minutes
  • Without these constraints, mathematically you could have solutions like x = -5, y = 10, which make sense mathematically but not in the context of the problem
  • These constraints ensure that your solution makes practical sense

Always include non-negativity constraints when variables represent countable items or measurable quantities that can't be negative.

Question: How do I check if my inequality correctly models the word problem?

Answer: Here's a systematic approach to verify your inequality:

  1. Check the direction: Does the inequality symbol match the constraint? "At most" means ≤, "at least" means ≥
  2. Test boundary values: If your constraint is x + y ≤ 50, test (25,25) which equals 50 and should satisfy the constraint
  3. Test impossible values: Test (30,30) which gives 60 > 50 and should violate the constraint
  4. Verify units: Make sure all terms have consistent units (dollars with dollars, pounds with pounds, etc.)

Also, read your inequality back in English to see if it matches the original constraint described in the problem.