Special Systems and Applications

4. Money and Cost Problems

Introduction

Money and cost problems are among the most practical applications of systems of linear equations. Every day, people and businesses make decisions involving prices, quantities, expenses, revenue, and profit.

For example:

• A student buys notebooks and pens but does not know the individual prices.

• A cinema sells adult and child tickets at different prices.

• Two companies offer different payment plans.

• A business wants to know how many products it must sell to cover its costs.

• A school sells two types of merchandise and needs to determine how many of each were sold.

• A customer compares two mobile phone plans to decide which offers better value.

Many of these situations involve two unknown quantities and two relationships. By representing the information using equations, we can construct a system of linear equations and solve it.

The goal is not simply to calculate numbers. We must also understand what the numbers represent, check whether they are realistic, and explain how the mathematical solution answers the original financial question.

Understanding Financial Relationships

Before solving money and cost problems, it is important to understand several common financial quantities.

Cost

Cost is the amount of money required to purchase, produce, or provide something.

Examples include the cost of buying supplies, manufacturing products, renting equipment, or operating a business.

A business may have two main types of costs:

• Fixed costs: Expenses that do not change with the number of products made within the model, such as monthly rent.

• Variable costs: Expenses that change with the number of products made, such as materials and packaging.

A common linear cost model is:

C(x) = mx + b

where:

• C(x) = total cost.

• x = number of products or units.

• m = variable cost per unit.

• b = fixed cost.

Revenue

Revenue is the total money received from selling goods or services before deducting costs.

For a product sold at a constant price:

Revenue = Selling price × Quantity sold

For example, selling 40 products at $15 each gives:

R = 15 × 40 = $600

Profit

Profit is the money remaining after costs are deducted from revenue.

Profit = Revenue − Cost

If a business earns $600 in revenue and has $450 in costs:

Profit = 600 − 450

Profit = $150

If costs exceed revenue, the business makes a loss.

Break-even point

The break-even point occurs when total revenue equals total cost.

Revenue = Cost

At this point:

Profit = $0

Finding the break-even point is one of the most important applications of systems of equations in business.

Identifying Variables in Financial Problems

A financial word problem often contains two unknown quantities.

The first step is defining them clearly.

For example:

"A customer purchases three notebooks and two pens for $17. Another customer purchases two notebooks and four pens for $18."

We need to determine the individual prices.

Define:

x = price of one notebook in dollars

y = price of one pen in dollars

These definitions allow us to write equations using the quantities purchased and the total amounts paid.

Translating financial statements

The most important skill is recognizing how the written information becomes mathematical relationships.

Constructing Systems from Purchase Problems

Suppose a shop sells notebooks and pens.

Three notebooks and two pens cost $17.

Two notebooks and four pens cost $18.

Step 1: Define variables

x = price of one notebook

y = price of one pen

Step 2: Write equations

First purchase:

3x + 2y = 17

Second purchase:

2x + 4y = 18

We now have a system of two equations with two unknowns.

Step 3: Solve using elimination

Multiply the first equation by 2:

6x + 4y = 34

Subtract the second equation:

(6x + 4y) − (2x + 4y) = 34 − 18

4x = 16

x = 4

Substitute into the first equation:

3(4) + 2y = 17

12 + 2y = 17

2y = 5

y = 2.50

Step 4: Interpret

One notebook costs $4.

One pen costs $2.50.

Step 5: Verify

First purchase:

3(4) + 2(2.50) = 12 + 5 = $17

Second purchase:

2(4) + 4(2.50) = 8 + 10 = $18

Both equations are satisfied.

Answer: Notebook = $4; pen = $2.50.

A General Method for Financial Systems

A reliable approach to financial word problems is:

1. Read the problem and identify the unknown quantities.

2. Define variables with appropriate units.

3. Translate the financial relationships into equations.

4. Solve the system using substitution, elimination, or graphing.

5. Interpret the values in the original context.

6. Verify that the solution satisfies every condition.

7. Check whether the result is financially and practically reasonable.

For example, if x represents the number of tickets sold, x should normally be a nonnegative whole number.

If x represents a price, a fractional value may be acceptable, but it should be expressed correctly in dollars and cents.

Worked Example: Ticket Sales

A school concert sells two types of tickets:

• Adult tickets cost $15 each.

• Student tickets cost $9 each.

A total of 140 tickets are sold.

The total revenue is $1,740.

How many adult and student tickets were sold?

Step 1: Define variables

a = number of adult tickets

s = number of student tickets

Step 2: Write the total ticket equation

a + s = 140

Step 3: Write the revenue equation

Adult ticket revenue:

15a

Student ticket revenue:

9s

Therefore:

15a + 9s = 1740

Step 4: Solve

From the first equation:

s = 140 − a

Substitute:

15a + 9(140 − a) = 1740

Expand:

15a + 1260 − 9a = 1740

Combine:

6a + 1260 = 1740

6a = 480

a = 80

Then:

s = 140 − 80

s = 60

Step 5: Interpret

The school sold:

• 80 adult tickets.

• 60 student tickets.

Step 6: Verify

Total tickets:

80 + 60 = 140

Total revenue:

15(80) + 9(60)

= 1200 + 540

= $1740

Answer: 80 adult tickets and 60 student tickets.

Worked Example: Comparing Two Payment Plans

A customer wants to compare two mobile phone plans.

Plan A charges a fixed monthly fee of $20 plus $5 per unit of additional usage.

Plan B charges a fixed monthly fee of $35 plus $2 per unit of additional usage.

At what usage level will the two plans cost the same?

Step 1: Define variables

x = number of additional usage units

y = total monthly cost in dollars

Step 2: Write the cost equations

Plan A:

y = 5x + 20

Plan B:

y = 2x + 35

Step 3: Solve the system

Set the equations equal:

5x + 20 = 2x + 35

Subtract 2x:

3x + 20 = 35

Subtract 20:

3x = 15

x = 5

Step 4: Find the cost

Substitute x = 5:

Plan A:

y = 5(5) + 20 = 45

Plan B:

y = 2(5) + 35 = 45

Step 5: Interpret

Both plans cost $45 when the customer uses five additional units.

For fewer than five units, Plan A is cheaper.

For more than five units, Plan B is cheaper.

Step 6: Verify

Both equations produce $45 at x = 5.

Answer: The plans have equal costs at five additional usage units, when each costs $45.

Graphical Interpretation of Payment Plans

A system of cost equations can be represented by two straight lines.

The point where the lines intersect shows the usage level at which the costs are equal.

This graph also illustrates the meaning of the slope and y-intercept.

• The slope represents the cost per additional usage unit.

• The y-intercept represents the fixed monthly fee.

• The intersection represents the break-even usage level between the two plans.

Worked Example: Taxi Fares

Two taxi companies charge different rates.

Company A charges a starting fee of $6 and $2 per kilometre.

Company B charges a starting fee of $3 and $2.50 per kilometre.

At what distance will the fares be equal?

Step 1: Define variables

d = distance travelled in kilometres

C = total fare in dollars

Step 2: Construct equations

Company A:

C = 2d + 6

Company B:

C = 2.50d + 3

Step 3: Solve

2d + 6 = 2.50d + 3

Subtract 2d:

6 = 0.50d + 3

Subtract 3:

3 = 0.50d

d = 6

Step 4: Calculate the fare

Company A:

C = 2(6) + 6 = 18

Company B:

C = 2.50(6) + 3 = 18

Step 5: Interpret

The taxi companies charge the same fare for a 6 km journey.

Both fares are $18.

For journeys shorter than 6 km, Company B is cheaper.

For journeys longer than 6 km, Company A is cheaper.

Answer: 6 km, at a fare of $18.

Worked Example: Business Break-Even Analysis

A small business manufactures reusable water bottles.

Its monthly fixed costs are $1,200.

Each bottle costs $8 to produce.

The selling price is $20 per bottle.

How many bottles must the business sell to break even?

Step 1: Define variables

x = number of bottles produced and sold

y = total amount in dollars

Step 2: Write the cost equation

Fixed costs:

$1,200

Variable cost per bottle:

$8

Therefore:

C(x) = 8x + 1200

Step 3: Write the revenue equation

Each bottle sells for $20.

Therefore:

R(x) = 20x

Step 4: Find the break-even point

At break-even:

C(x) = R(x)

8x + 1200 = 20x

1200 = 12x

x = 100

Step 5: Calculate cost and revenue

Cost:

C(100) = 8(100) + 1200

C(100) = 2000

Revenue:

R(100) = 20(100)

R(100) = 2000

Step 6: Interpret

The business must sell 100 bottles to break even.

At this quantity:

• Total cost = $2,000.

• Total revenue = $2,000.

• Profit = $0.

Selling fewer than 100 bottles produces a loss.

Selling more than 100 bottles produces a profit under this model.

Answer: The break-even point is 100 bottles, corresponding to $2,000 in cost and revenue.

Understanding Profit Functions

Profit can be represented mathematically.

P(x) = R(x) − C(x)

For the water bottle business:

R(x) = 20x

C(x) = 8x + 1200

Therefore:

P(x) = 20x − (8x + 1200)

Simplify:

P(x) = 12x − 1200

This equation can be used to predict profit at different sales quantities.

Example: Selling 150 bottles

P(150) = 12(150) − 1200

= 1800 − 1200

= $600

Example: Selling 60 bottles

P(60) = 12(60) − 1200

= 720 − 1200

= −$480

A negative profit represents a loss.

Example: Target profit

How many bottles must the business sell to earn $1,200 profit?

Set:

12x − 1200 = 1200

12x = 2400

x = 200

The business must sell 200 bottles to earn $1,200 profit.

These calculations assume every bottle produced is sold and that unit costs and selling prices remain constant.

Worked Example: Finding Unknown Prices

A school canteen sells sandwiches and drinks.

Three sandwiches and two drinks cost $21.

Two sandwiches and five drinks cost $24.

Find the price of one sandwich and one drink.

Step 1: Define variables

s = price of one sandwich in dollars

d = price of one drink in dollars

Step 2: Construct equations

First purchase:

3s + 2d = 21

Second purchase:

2s + 5d = 24

Step 3: Solve using elimination

Multiply the first equation by 5:

15s + 10d = 105

Multiply the second equation by 2:

4s + 10d = 48

Subtract:

11s = 57

s = 57/11 ≈ 5.18

Substitute into the first equation:

3(57/11) + 2d = 21

171/11 + 2d = 231/11

2d = 60/11

d = 30/11 ≈ 2.73

Step 4: Interpret

The exact algebraic prices are:

• Sandwich = $57/11 ≈ $5.18.

• Drink = $30/11 ≈ $2.73.

However, these exact values are not whole numbers of cents.

If the canteen charges only whole cents, the original totals do not produce a valid exact pricing solution under that restriction.

This illustrates why financial solutions must be checked for practical reasonableness.

A revised example

Suppose the second purchase instead costs $25.

Then:

3s + 2d = 21

2s + 5d = 25

Multiply the first equation by 5:

15s + 10d = 105

Multiply the second equation by 2:

4s + 10d = 50

Subtract:

11s = 55

s = 5

Substitute:

3(5) + 2d = 21

15 + 2d = 21

d = 3

Verification

First purchase:

3(5) + 2(3) = 21

Second purchase:

2(5) + 5(3) = 25

For the revised totals, a sandwich costs $5 and a drink costs $3.

Worked Example: Finding Numbers of Coins

A student has 30 coins consisting only of 20-cent and 50-cent coins.

The total value is $9.00.

How many of each coin does the student have?

Step 1: Define variables

x = number of 20-cent coins

y = number of 50-cent coins

Step 2: Construct equations

Total coins:

x + y = 30

Total value:

0.20x + 0.50y = 9

Step 3: Solve

From the first equation:

x = 30 − y

Substitute:

0.20(30 − y) + 0.50y = 9

6 − 0.20y + 0.50y = 9

6 + 0.30y = 9

0.30y = 3

y = 10

Then:

x = 30 − 10

x = 20

Step 4: Interpret

The student has:

• 20 twenty-cent coins.

• 10 fifty-cent coins.

Step 5: Verify

Total number:

20 + 10 = 30 coins

Total value:

0.20(20) + 0.50(10)

= 4 + 5

= $9

Answer: 20 twenty-cent coins and 10 fifty-cent coins.

Worked Example: Finding Two Unknown Quantities from Revenue

A school sells T-shirts and caps at a fundraising event.

T-shirts cost $18 each.

Caps cost $12 each.

A total of 85 items are sold, generating $1,320.

How many T-shirts and caps were sold?

Step 1: Define variables

t = number of T-shirts sold

c = number of caps sold

Step 2: Construct equations

Total items:

t + c = 85

Total revenue:

18t + 12c = 1320

Step 3: Solve

Substitute c = 85 − t:

18t + 12(85 − t) = 1320

18t + 1020 − 12t = 1320

6t = 300

t = 50

Then:

c = 85 − 50

c = 35

Step 4: Verify

Total items:

50 + 35 = 85

Total revenue:

18(50) + 12(35)

= 900 + 420

= $1320

Answer: 50 T-shirts and 35 caps were sold.

Worked Example: Comparing Production Costs

Two companies manufacture the same product using different production methods.

Company A has fixed costs of $800 and variable costs of $12 per unit.

Company B has fixed costs of $1,400 and variable costs of $8 per unit.

At what production quantity will their costs be equal?

Step 1: Define variables

x = number of products manufactured

C = total production cost

Step 2: Construct equations

Company A:

C = 12x + 800

Company B:

C = 8x + 1400

Step 3: Solve

12x + 800 = 8x + 1400

4x = 600

x = 150

Step 4: Calculate the equal cost

Company A:

C = 12(150) + 800

C = 2600

Company B:

C = 8(150) + 1400

C = 2600

Step 5: Interpret

At 150 products, both methods cost $2,600.

For fewer than 150 products, Company A has lower costs.

For more than 150 products, Company B has lower costs.

Answer: 150 products, at a total cost of $2,600.

This example shows why a business might choose a production method with higher fixed costs but lower variable costs when production quantities are large.

Solving Financial Systems Using Graphs

Graphs provide a useful way to compare financial relationships.

Consider:

Company A:

C = 12x + 800

Company B:

C = 8x + 1400