Inequalities and Budget Limits
Learning objectives
After this section you will be able to:
- Translate budget and minimum requirements into inequalities.
- Solve inequalities, reversing the sign when required.
- Interpret solution sets and whole-number restrictions.
Real-life situation
Planning a student workshop
A hypothetical workshop has a $20 setup charge and an $8 supply cost per participant. The organizers have at most $100 to spend. The budget permits a range of participant counts rather than just one number.
| Participants q | Cost ($) | Within $100? |
|---|---|---|
| 0 | 20 | Yes |
| 5 | 60 | Yes |
| 10 | 100 | Yes |
The concept
An inequality compares quantities that need not be equal. The symbols < and > mean strictly less than and strictly greater than. The symbols ≤ and ≥ include equality. Translate 'at most' as ≤ and 'at least' as ≥.
Solve much as you solve an equation: add or subtract the same quantity on both sides. Multiplying or dividing both sides by a positive number preserves the inequality direction. Multiplying or dividing by a negative number reverses it because the order of numbers reverses.
A solution set contains every value making the original statement true. On a number line, use an open endpoint for < or > and a closed endpoint for ≤ or ≥. Shade toward smaller or larger values as appropriate.
Context can narrow the algebraic solution set. A budget may give q ≤ 13.75, but participant counts must be nonnegative whole numbers. The largest feasible count is therefore 13, and checking the next integer confirms why rounding to 14 would violate the limit.
Inequalities and Budget Limits: solution map
- 1
Identify the question
How many participants can attend if setup costs $20, supplies cost $8 each, and the budget is $100?
- 2
Calculation 1
Define q as a nonnegative whole-number participant count.
- 3
Calculation 2
Write 20 + 8q ≤ 100.
- 4
Calculation 3
Subtract 20: 8q ≤ 80.
- 5
Calculation 4
Divide by positive 8: q ≤ 10.
- 6
Calculation 5
Check the endpoint: 20 + 8 × 10 = 100.
- 7
Interpret and check
The feasible counts are 0 through 10 participants.
Money Words
Plain-language words you will use in this lesson — learn them before we reach the formulas.
budget
A plan that shows expected income and how it will be spent.
loan term
How long a loan lasts before it must be fully repaid.
total repayment cost
The full amount paid over a loan's life — principal plus all interest and fees.
Key terms
- Inequality
- A statement comparing quantities using <, >, ≤, or ≥.
- Solution set
- All values that make a statement true.
- Boundary
- A value separating included and excluded parts of a solution set.
- Constraint
- A limit or requirement that a plan must satisfy.
- Feasible value
- A value satisfying every relevant constraint.
Formulas & rules
Budget constraint
F + v × q ≤ B
B is the maximum budget.
Positive per-unit cost
q ≤ (B − F) ÷ v
For v > 0; also apply nonnegative and whole-number restrictions.
Negative multiplication rule
If a < b and c < 0, then ac > bc
Worked examples
How many participants can attend if setup costs $20, supplies cost $8 each, and the budget is $100?
Solution
- 1Define q as a nonnegative whole-number participant count.
- 2Write 20 + 8q ≤ 100.
- 3Subtract 20: 8q ≤ 80.
- 4Divide by positive 8: q ≤ 10.
- 5Check the endpoint: 20 + 8 × 10 = 100.
Answer: The feasible counts are 0 through 10 participants.
Solve −3x ≤ −12 and describe the number-line graph.
Solution
- 1Divide both sides by −3.
- 2Reverse ≤ to ≥ because the divisor is negative.
- 3Obtain x ≥ 4.
- 4Check x = 4: −12 ≤ −12 is true; check x = 5: −15 ≤ −12 is true.
- 5Use a closed dot at 4 and shade to the right.
Answer: x ≥ 4.
Practice exercises 24.1
- Solve 15 + 9q ≤ 60 for nonnegative whole-number q.
- A project costs $35 plus $12 per kit. With a $200 budget, what is the maximum whole-number kit count?
- Solve 25 + 15h ≥ 100.
- Solve −4x + 10 > 22.
- Solve 14 + 6q < 50 for nonnegative whole-number q.
Recap quiz 24.1
Section 24.1 recap
- Translate the boundary carefully: 'at most' includes equality; 'less than' does not.
- Reverse the sign after negative multiplication or division, then apply context restrictions.
Answer key with explanations: Section 24.1
Practice exercises
- 9q ≤ 45, so q ≤ 5. The feasible whole-number counts are 0 through 5.
- 35 + 12q ≤ 200 gives q ≤ 165 ÷ 12 = 13.75. Use at most 13 kits: cost $191.00; 14 would cost $203.00.
- Subtract 25: 15h ≥ 75. Divide by 15: h ≥ 5.
- Subtract 10: −4x > 12. Divide by −4 and reverse the sign: x < −3. Use an open dot at −3 and shade left.
- 6q < 36 gives q < 6. Since equality is excluded, the allowed counts are 0 through 5.
Recap quiz
- x ≤ 6: 5x ≤ 30, so x ≤ 6.
- When dividing by a negative number: Multiplying or dividing by a negative reverses numerical order.
- At least 10: At least includes 10 and every larger allowed value.