WealthWiseGuys Practical Mathematics

Chapter 24

Inequalities, Systems of Equations, and Decision-Making

Describe limits, compare alternatives, and choose feasible plans

Budgets create limits, competing plans create comparison points, and limited resources create trade-offs. Inequalities and systems help you describe those situations without guessing. All people, businesses, prices, and earnings are hypothetical; round money to the nearest cent when needed.

24.1

Inequalities and Budget Limits

Learning objectives

After this section you will be able to:

  1. Translate budget and minimum requirements into inequalities.
  2. Solve inequalities, reversing the sign when required.
  3. 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 qCost ($)Within $100?
020Yes
560Yes
10100Yes
Figure 24.1 Hypothetical workshop budget

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. 1

    Identify the question

    How many participants can attend if setup costs $20, supplies cost $8 each, and the budget is $100?

  2. 2

    Calculation 1

    Define q as a nonnegative whole-number participant count.

  3. 3

    Calculation 2

    Write 20 + 8q ≤ 100.

  4. 4

    Calculation 3

    Subtract 20: 8q ≤ 80.

  5. 5

    Calculation 4

    Divide by positive 8: q ≤ 10.

  6. 6

    Calculation 5

    Check the endpoint: 20 + 8 × 10 = 100.

  7. 7

    Interpret and check

    The feasible counts are 0 through 10 participants.

Figure 24.1.V Inequalities and Budget Limits: solution map. Hypothetical teaching data.

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

Example 1Maximum workshop attendance

How many participants can attend if setup costs $20, supplies cost $8 each, and the budget is $100?

Solution

  1. 1Define q as a nonnegative whole-number participant count.
  2. 2Write 20 + 8q ≤ 100.
  3. 3Subtract 20: 8q ≤ 80.
  4. 4Divide by positive 8: q ≤ 10.
  5. 5Check the endpoint: 20 + 8 × 10 = 100.

Answer: The feasible counts are 0 through 10 participants.

Example 2Reverse the inequality

Solve −3x ≤ −12 and describe the number-line graph.

Solution

  1. 1Divide both sides by −3.
  2. 2Reverse ≤ to ≥ because the divisor is negative.
  3. 3Obtain x ≥ 4.
  4. 4Check x = 4: −12 ≤ −12 is true; check x = 5: −15 ≤ −12 is true.
  5. 5Use a closed dot at 4 and shade to the right.

Answer: x ≥ 4.

Practice exercises 24.1

  1. Solve 15 + 9q ≤ 60 for nonnegative whole-number q.
  2. A project costs $35 plus $12 per kit. With a $200 budget, what is the maximum whole-number kit count?
  3. Solve 25 + 15h ≥ 100.
  4. Solve −4x + 10 > 22.
  5. Solve 14 + 6q < 50 for nonnegative whole-number q.

Recap quiz 24.1

1. Solve 10 + 5x ≤ 40.
2. When must an inequality sign reverse?
3. Which phrase means x ≥ 10?

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

  1. 9q ≤ 45, so q ≤ 5. The feasible whole-number counts are 0 through 5.
  2. 35 + 12q ≤ 200 gives q ≤ 165 ÷ 12 = 13.75. Use at most 13 kits: cost $191.00; 14 would cost $203.00.
  3. Subtract 25: 15h ≥ 75. Divide by 15: h ≥ 5.
  4. Subtract 10: −4x > 12. Divide by −4 and reverse the sign: x < −3. Use an open dot at −3 and shade left.
  5. 6q < 36 gives q < 6. Since equality is excluded, the allowed counts are 0 through 5.

Recap quiz

  1. x ≤ 6: 5x ≤ 30, so x ≤ 6.
  2. When dividing by a negative number: Multiplying or dividing by a negative reverses numerical order.
  3. At least 10: At least includes 10 and every larger allowed value.
24.2

Solving Systems and Comparing Plans

Learning objectives

After this section you will be able to:

  1. Solve two linear equations by substitution or elimination.
  2. Interpret an intersection as a shared solution.
  3. Recognize systems with no solution or infinitely many solutions.

Real-life situation

Two hypothetical data plans

Plan A costs $12 plus $3 per data unit. Plan B costs $24 plus $1 per data unit. Comparing both equations identifies when the plans cost the same and which is cheaper on either side.

Data units gPlan A ($)Plan B ($)
01224
42428
63030
Figure 24.2 Hypothetical plan charges

The concept

A system asks for values satisfying two or more equations at once. With two variables, a solution is an ordered pair. Graphically, it lies on both lines, so an intersection represents a shared solution.

Substitution replaces a variable with an equal expression from another equation. When both plans are written as a total cost, setting their cost expressions equal is substitution: the same output must equal both expressions.

Elimination adds or subtracts equations to remove a variable. First multiply an equation if necessary to create equal or opposite coefficients. After finding one variable, substitute back to find the other, and check both original equations.

Two distinct nonparallel lines meet once. Parallel distinct lines have no common solution, while two equations describing the same line have infinitely many solutions. An algebraic result such as 0 = 5 signals inconsistency; a true identity such as 0 = 0 signals dependence when the original equations describe the same line.

An intersection is not automatically the best choice. Compare prices before and after it and consider restrictions such as capacity, included features, and allowed inputs. A mathematical solution outside the permitted domain may not be usable.

Solving Systems and Comparing Plans: solution map

  1. 1

    Identify the question

    At what usage do Plans A and B cost the same?

  2. 2

    Calculation 1

    Set costs equal: 12 + 3g = 24 + g.

  3. 3

    Calculation 2

    Subtract g and 12: 2g = 12.

  4. 4

    Calculation 3

    Divide by 2: g = 6.

  5. 5

    Calculation 4

    Substitute: A = 12 + 3 × 6 = 30 and B = 24 + 6 = 30.

  6. 6

    Calculation 5

    Because A starts lower but rises faster, A is cheaper below 6 units and B is cheaper above 6 units.

  7. 7

    Interpret and check

    The plans meet at (6, 30): 6 data units and $30.00.

Figure 24.2.V Solving Systems and Comparing Plans: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • down payment

    Money paid upfront when financing a purchase, which reduces the amount borrowed.

  • monthly payment

    The amount paid each month toward a loan or installment plan.

  • interest

    The cost of borrowing money, or the earnings on money saved or invested.

Key terms

System of equations
Equations required to hold at the same time.
Substitution
Replacing a quantity with an expression equal to it.
Elimination
Combining equations to remove a variable.
Intersection
A point shared by graphs.
Inconsistent system
A system with no solution.
Dependent equations
Equations that describe the same relationship.

Formulas & rules

  • Equal-cost comparison

    F₁ + r₁x = F₂ + r₂x

  • Comparison input

    x = (F₂ − F₁) ÷ (r₁ − r₂)

    Requires different rates.

  • Two-category total

    x + y = n

Worked examples

Example 1Compare two plans

At what usage do Plans A and B cost the same?

Solution

  1. 1Set costs equal: 12 + 3g = 24 + g.
  2. 2Subtract g and 12: 2g = 12.
  3. 3Divide by 2: g = 6.
  4. 4Substitute: A = 12 + 3 × 6 = 30 and B = 24 + 6 = 30.
  5. 5Because A starts lower but rises faster, A is cheaper below 6 units and B is cheaper above 6 units.

Answer: The plans meet at (6, 30): 6 data units and $30.00.

Example 2Eliminate a ticket category

A hypothetical event sells 10 tickets: student tickets cost $2 and adult tickets cost $4. Revenue is $28. How many of each sold?

Solution

  1. 1Let s be student tickets and a be adult tickets: s + a = 10 and 2s + 4a = 28.
  2. 2Multiply the count equation by 2: 2s + 2a = 20.
  3. 3Subtract it from the revenue equation: 2a = 8.
  4. 4Divide by 2: a = 4; then s = 10 − 4 = 6.
  5. 5Check: 6 + 4 = 10 and 2 × 6 + 4 × 4 = 28.

Answer: 6 student tickets and 4 adult tickets.

Practice exercises 24.2

  1. Find the equal-cost input for A = 8 + 4x and B = 20 + 2x.
  2. Solve x + y = 10 and 4x + 7y = 70.
  3. Do y = 3x + 4 and y = 3x + 9 intersect?
  4. How many solutions do y = 2x + 1 and 2y = 4x + 2 have?
  5. Find the intersection of y = 5x + 1 and y = 3x + 5.

Recap quiz 24.2

1. Which point satisfies y = 3x + 4?
2. A system's solution must satisfy:
3. Different lines with the same slope have:

Section 24.2 recap

  • Solve by substitution or elimination, then check both equations.
  • An intersection can be a comparison threshold rather than a universally best choice.
Answer key with explanations: Section 24.2

Practice exercises

  1. 8 + 4x = 20 + 2x gives 2x = 12, so x = 6. Both costs are $32.00.
  2. Multiply the first equation by 4: 4x + 4y = 40. Subtract to get 3y = 30, so y = 10 and x = 0. Check: 0 + 70 = 70.
  3. Setting them equal gives 3x + 4 = 3x + 9, or 4 = 9. This is false, so the parallel lines have no solution.
  4. Divide the second equation by 2 to obtain y = 2x + 1. Both describe the same line, so there are infinitely many shared points.
  5. 5x + 1 = 3x + 5 gives 2x = 4, so x = 2. Then y = 11; both equations give 11.

Recap quiz

  1. (2, 10): At x = 2, y = 3 × 2 + 4 = 10.
  2. Every equation in the system: A shared solution makes all original equations true.
  3. No intersection: Different intercepts with equal slopes produce parallel lines.
24.3

Feasible Regions and Resource Decisions

Learning objectives

After this section you will be able to:

  1. Represent multiple resource limits with inequalities.
  2. Test proposed plans against every constraint.
  3. Compare an objective at feasible corner points.

Real-life situation

A student print-shop decision

A hypothetical student shop makes x poster packs and y card packs. It can prepare at most 8 packs total and has 12 machine-time units; poster packs use 2 units and card packs use 1. Modeled contribution is $15 per poster pack and $10 per card pack.

Poster packs xCard packs yTotal packsMachine unitsContribution ($)
00000
066660
44812100
6061290
Figure 24.3 Candidate plans and modeled contribution

The concept

A practical plan may have several limits at once. Here x + y ≤ 8 limits total packs, while 2x + y ≤ 12 limits machine time. Also x ≥ 0 and y ≥ 0, and pack counts must be whole numbers.

To graph an inequality in two variables, first draw its boundary line. Use a solid line when equality is allowed and a dashed line for a strict inequality. Test a point not on the line to determine which side to shade. The overlap of all permitted regions is the feasible region.

The objective is the quantity you want to maximize or minimize. Here contribution is Z = 15x + 10y. Contribution is not necessarily final profit: any fixed expenses not included in these amounts would still need to be subtracted.

For a linear objective over a nonempty bounded polygonal feasible region, a maximum occurs at a corner; sometimes an entire edge ties. Find corners by intersecting boundary lines and axes, discard infeasible intersections, and evaluate the objective at the remaining corners.

Whole-number restrictions require care. A fractional best corner may be unusable, and rounding it can violate constraints or miss the best integer plan. In this example the best continuous corner already has whole-number coordinates, so it also solves the whole-pack problem.

Feasible Regions and Resource Decisions: solution map

  1. 1

    Identify the question

    Find the intersection of x + y = 8 and 2x + y = 12, and check feasibility.

  2. 2

    Calculation 1

    Subtract the first equation from the second: x = 4.

  3. 3

    Calculation 2

    Substitute into x + y = 8: y = 4.

  4. 4

    Calculation 3

    Check total packs: 4 + 4 = 8.

  5. 5

    Calculation 4

    Check machine time: 2 × 4 + 4 = 12.

  6. 6

    Calculation 5

    Both counts are nonnegative whole numbers, so (4, 4) is feasible.

  7. 7

    Interpret and check

    The boundaries intersect at 4 poster packs and 4 card packs.

Figure 24.3.V Feasible Regions and Resource Decisions: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • APR

    Annual percentage rate — the yearly cost of borrowing, including interest and some fees, shown as a percent.

  • collateral

    Property a borrower promises to a lender if the loan is not repaid, such as a car.

  • profit margin

    Profit shown as a percent of revenue; a 10% margin means $10 of profit per $100 of sales.

Key terms

Feasible region
All points satisfying every constraint.
Boundary line
The equality line associated with a linear inequality.
Objective function
A formula for the quantity to maximize or minimize.
Corner point
A vertex where boundary segments of a feasible region meet.
Optimal solution
A feasible solution with the best objective value.

Formulas & rules

  • Pack limit

    x + y ≤ 8

  • Machine-time limit

    2x + y ≤ 12

  • Contribution objective

    Z = 15x + 10y

Worked examples

Example 1Find the shared boundary

Find the intersection of x + y = 8 and 2x + y = 12, and check feasibility.

Solution

  1. 1Subtract the first equation from the second: x = 4.
  2. 2Substitute into x + y = 8: y = 4.
  3. 3Check total packs: 4 + 4 = 8.
  4. 4Check machine time: 2 × 4 + 4 = 12.
  5. 5Both counts are nonnegative whole numbers, so (4, 4) is feasible.

Answer: The boundaries intersect at 4 poster packs and 4 card packs.

Example 2Compare all corners

Find the highest contribution under x + y ≤ 8, 2x + y ≤ 12, x ≥ 0, and y ≥ 0.

Solution

  1. 1The axis corners are (0, 0), (0, 8), and (6, 0); the boundary intersection is (4, 4).
  2. 2At (0, 0), Z = 0.
  3. 3At (0, 8), Z = 15 × 0 + 10 × 8 = 80.
  4. 4At (6, 0), Z = 15 × 6 = 90.
  5. 5At (4, 4), Z = 60 + 40 = 100, the greatest corner value.

Answer: Make 4 poster packs and 4 card packs for a maximum modeled contribution of $100.00.

Practice exercises 24.3

  1. Is (3, 5) feasible for the shop?
  2. Is (5, 3) feasible?
  3. Compare contribution at (3, 5) and (4, 4).
  4. For y = 0, what is the largest feasible x?
  5. A student says (0, 6) is a corner because it is listed in Figure 24.3. Correct the claim.

Recap quiz 24.3

1. Which plan violates machine time?
2. At (5, 2), how many machine-time units are used?
3. What does 'feasible' mean?

Section 24.3 recap

  • Test every constraint; satisfying one limit is not enough.
  • Compare feasible corners for a linear objective, and handle integer restrictions separately.
Answer key with explanations: Section 24.3

Practice exercises

  1. Total packs: 3 + 5 = 8 ≤ 8. Machine time: 2 × 3 + 5 = 11 ≤ 12. Both counts are nonnegative integers, so yes.
  2. Total packs equal 8, but machine time is 2 × 5 + 3 = 13 > 12. It is not feasible.
  3. At (3, 5), Z = 45 + 50 = $95.00. At (4, 4), Z = 60 + 40 = $100.00, which is $5.00 higher.
  4. The constraints give x ≤ 8 and 2x ≤ 12, so x ≤ 6. The tighter limit permits 6 poster packs.
  5. The point is feasible, but it is not a corner. On the y-axis, the limits give y ≤ 8 and y ≤ 12, so the upper corner is (0, 8). A candidate table need not list only corners.

Recap quiz

  1. (5, 3): At (5, 3), 2x + y = 13, exceeding 12.
  2. 12: 2 × 5 + 2 = 12; total packs are 7, so both limits are satisfied.
  3. Satisfies every constraint: A feasible plan obeys all limits; it need not be optimal or use all resources.

Chapter 24 Organizer and Review

Chapter Organizer

Every section at a glance: key formulas, key terms, and the main idea.

  • 24.1 Inequalities and Budget Limits

    Key formulas

    • Budget constraint: F + v × q ≤ B
    • Positive per-unit cost: q ≤ (B − F) ÷ v
    • Negative multiplication rule: If a < b and c < 0, then ac > bc

    Key terms

    Inequality, Solution set, Boundary, Constraint, Feasible value

    Remember

    Translate the boundary carefully: 'at most' includes equality; 'less than' does not.

  • 24.2 Solving Systems and Comparing Plans

    Key formulas

    • Equal-cost comparison: F₁ + r₁x = F₂ + r₂x
    • Comparison input: x = (F₂ − F₁) ÷ (r₁ − r₂)
    • Two-category total: x + y = n

    Key terms

    System of equations, Substitution, Elimination, Intersection, Inconsistent system, Dependent equations

    Remember

    Solve by substitution or elimination, then check both equations.

  • 24.3 Feasible Regions and Resource Decisions

    Key formulas

    • Pack limit: x + y ≤ 8
    • Machine-time limit: 2x + y ≤ 12
    • Contribution objective: Z = 15x + 10y

    Key terms

    Feasible region, Boundary line, Objective function, Corner point, Optimal solution

    Remember

    Test every constraint; satisfying one limit is not enough.

Cumulative Review

Mixed practice from this chapter and the two chapters before it.

  1. Section 24.1

    Solve 14 + 6q < 50 for nonnegative whole-number q.

    Show answer

    6q < 36 gives q < 6. Since equality is excluded, the allowed counts are 0 through 5.

  2. Section 24.2

    Find the intersection of y = 5x + 1 and y = 3x + 5.

    Show answer

    5x + 1 = 3x + 5 gives 2x = 4, so x = 2. Then y = 11; both equations give 11.

  3. Section 24.3

    A student says (0, 6) is a corner because it is listed in Figure 24.3. Correct the claim.

    Show answer

    The point is feasible, but it is not a corner. On the y-axis, the limits give y ≤ 8 and y ≤ 12, so the upper corner is (0, 8). A candidate table need not list only corners.

  4. Review of Section 22.5

    Find wraps’ unit share and revenue share.

    Show answer

    Unit share = 80 ÷ 400 × 100% = 20.00%; revenue share = 480 ÷ 2,040 × 100% ≈ 23.53%.

  5. Review of Section 23.3

    Find the stand's profit at 40 items.

    Show answer

    P = (12 − 8) × 40 − 80 = 160 − 80 = $80.00.

All prices, stores, people, and accounts in this chapter are hypothetical and for teaching only. Product references are generic categories; no brand endorses or is endorsed by this textbook.