Applications and Mathematical Modelling

1. Business and Economics

Introduction

Businesses make decisions every day about prices, production, costs, sales, and profits. These decisions often involve several unknown quantities that are connected by mathematical relationships.

For example, a business owner may need to determine how many products must be sold to cover operating costs, compare two production methods, or calculate the quantities of different products sold from total revenue information.

Systems of equations provide a powerful way to represent and solve these problems.

By combining information about costs, revenue, and quantities, businesses can make informed decisions about pricing, production, and profitability.

Systems of equations are also used in economics to study supply and demand, market equilibrium, resource allocation, and financial planning.

In this topic, we will explore how to construct mathematical models of business situations, solve the resulting systems, and interpret the solutions in practical terms.

Why Are Systems of Equations Useful in Business?

A business rarely makes decisions using only one quantity.

For example, a company may need to consider:

• The number of products manufactured.

• The cost of producing each product.

• The selling price of each product.

• The fixed costs of operating the business.

• The revenue earned from sales.

• The profit or loss generated.

These quantities are often related.

If a company sells more products, its revenue generally increases. However, its production costs may also increase.

The business must understand how these relationships affect profitability.

Example: A Small Business

A company manufactures water bottles.

Each bottle sells for $15.

The variable production cost is $6 per bottle.

The company also pays $900 in fixed operating costs.

If x bottles are produced and sold, then:

Revenue:

R = 15x

Total cost:

C = 900 + 6x

Profit:

P = R − C

Therefore:

P = 15x − (900 + 6x)

P = 9x − 900

This equation shows how profit depends on the number of bottles sold.

If the company sells 100 bottles:

P = 9(100) − 900

P = 0

The business breaks even.

If the company sells 150 bottles:

P = 9(150) − 900

P = 450

The business earns a profit of $450.

If the company sells only 50 bottles:

P = 9(50) − 900

P = −450

The business makes a loss of $450.

These calculations illustrate how mathematical models can support business decisions.

Understanding Costs, Revenue, and Profit

Before solving business problems, we need to understand three important quantities.

Cost

Cost is the money a business spends to produce goods or provide services.

Costs are commonly divided into two categories.

Fixed costs do not change with the number of units produced within the relevant operating range.

Examples include:

• Rent.

• Insurance.

• Salaried administrative staff.

• Equipment leases.

• Software subscriptions.

Variable costs change with production or sales volume.

Examples include:

• Raw materials.

• Packaging.

• Production supplies.

• Per-unit shipping costs.

• Sales commissions.

Total Cost Formula

For a simple linear cost model:

C = F + vx

where:

• C = total cost.

• F = fixed cost.

• v = variable cost per unit.

• x = number of units produced.

For example, if fixed costs are $1,200 and variable costs are $8 per unit:

C = 1200 + 8x

Revenue

Revenue is the money received from selling goods or services.

For a constant selling price:

R = px

where:

• R = total revenue.

• p = selling price per unit.

• x = number of units sold.

If a product sells for $20:

R = 20x

Profit

Profit is the amount remaining after costs are subtracted from revenue.

P = R − C

A positive profit indicates that revenue exceeds cost.

A negative profit indicates a loss.

A profit of zero indicates the business is breaking even.

Summary

Understanding Break-Even Points

The break-even point is the production or sales level at which total revenue equals total cost.

At break-even:

R = C

This means:

Profit = 0

Businesses use break-even analysis to determine the minimum sales volume required to cover costs under a particular model.

Worked Example: Break-Even Analysis

A company sells reusable bottles for $18 each.

The variable cost is $8 per bottle.

Fixed costs are $1,500.

Find the break-even quantity.

Step 1: Define the variable

x = number of bottles sold

Step 2: Write the revenue equation

R = 18x

Step 3: Write the cost equation

C = 1500 + 8x

Step 4: Set revenue equal to cost

18x = 1500 + 8x

Subtract 8x:

10x = 1500

x = 150

Step 5: Interpret

The company must sell 150 bottles to break even.

Step 6: Verify

Revenue:

R = 18(150) = $2700

Cost:

C = 1500 + 8(150)

C = 1500 + 1200

C = $2700

Revenue equals cost.

Answer: The break-even point is 150 bottles, with revenue and cost both equal to $2,700.

Graphing Break-Even Problems

Break-even analysis can be represented using two straight lines.

One line represents revenue.

The other represents total cost.

The point where the lines intersect is the break-even point.

Interpreting the graph

Before the intersection:

Cost is greater than revenue.

The business operates at a loss.

At the intersection:

Revenue equals cost.

The business breaks even.

After the intersection:

Revenue is greater than cost.

The business earns a profit.

Important assumptions

This linear model assumes that:

• The selling price remains constant.

• Variable cost per unit remains constant.

• Fixed costs remain unchanged.

• All produced units are sold.

• The business has sufficient production capacity.

Real businesses may experience changing prices, discounts, additional costs, and production limits.

Therefore, the model is useful but simplified.

Application: Comparing Two Business Plans

A company is considering two ways to manufacture a product.

Plan A

Fixed costs: $2,000

Variable cost: $5 per unit

Plan B

Fixed costs: $800

Variable cost: $9 per unit

The company wants to know which plan is cheaper at different production levels.

Step 1: Define the variable

x = number of units manufactured

Step 2: Construct the cost equations

Plan A:

Cₐ = 2000 + 5x

Plan B:

Cᵦ = 800 + 9x

Step 3: Find the intersection

At the intersection, both plans have the same cost.

2000 + 5x = 800 + 9x

1200 = 4x

x = 300

Step 4: Calculate the cost

Plan A:

Cₐ = 2000 + 5(300)

Cₐ = $3500

Plan B:

Cᵦ = 800 + 9(300)

Cᵦ = $3500

The two plans cost the same when 300 units are manufactured.

Step 5: Compare different production levels

Business conclusion

Plan B is cheaper for fewer than 300 units because it has lower fixed costs.

Plan A is cheaper for more than 300 units because it has a lower variable cost per unit.

The decision depends on expected production volume.

This example demonstrates how solving a system can identify the point at which one business strategy becomes more economical than another.

Application: Finding Fixed and Variable Costs

A manufacturer records the following production costs:

• Producing 100 units costs $1,400.

• Producing 250 units costs $2,600.

Assuming a linear cost model, determine the fixed cost and variable cost per unit.

Step 1: Define variables

F = fixed cost, in dollars

v = variable cost per unit, in dollars

The model is:

C = F + vx

Step 2: Construct equations

For 100 units:

F + 100v = 1400

For 250 units:

F + 250v = 2600

Step 3: Eliminate F

Subtract the first equation from the second:

150v = 1200

v = 8

Step 4: Find F

F + 100(8) = 1400

F + 800 = 1400

F = 600

Step 5: Interpret

Fixed cost = $600

Variable cost = $8 per unit

Therefore:

C = 600 + 8x

Step 6: Verify

At 100 units:

C = 600 + 8(100) = $1400

At 250 units:

C = 600 + 8(250) = $2600

Both values match the original data.

Business conclusion

The company spends $600 regardless of production volume within the model's operating range, plus $8 for each unit manufactured.

This information can help the business forecast costs at other production levels.

Application: Determining Product Prices from Sales

A school shop sells notebooks and pens.

On Monday, it sells 30 notebooks and 20 pens for $190.

On Tuesday, it sells 20 notebooks and 40 pens for $200.

Find the selling price of each item.

Step 1: Define variables

n = selling price of one notebook

p = selling price of one pen

Step 2: Construct equations

Monday:

30n + 20p = 190

Tuesday:

20n + 40p = 200

Step 3: Simplify

Divide the first equation by 10:

3n + 2p = 19

Divide the second equation by 20:

n + 2p = 10

Step 4: Eliminate p

Subtract the second equation from the first:

2n = 9

n = 4.50

Step 5: Find p

4.50 + 2p = 10

2p = 5.50

p = 2.75

Step 6: Verify

Monday:

30(4.50) + 20(2.75)

= 135 + 55 = $190

Tuesday:

20(4.50) + 40(2.75)

= 90 + 110 = $200

Answer: Notebooks sell for $4.50 and pens sell for $2.75.

Economic interpretation

The system determines the prices that are consistent with both days' sales records, assuming the prices were unchanged and the recorded revenue came only from these items.

Application: Three Products and Total Revenue

A company sells three products: A, B, and C.

Their selling prices are:

• Product A: $10

• Product B: $15

• Product C: $20

During one day, the company sells 100 products and earns $1,500.

It sells 20 more units of Product A than Product C.

Find the quantity of each product sold.

Step 1: Define variables

a = number of Product A units

b = number of Product B units

c = number of Product C units

Step 2: Construct equations

Total units:

a + b + c = 100

Revenue:

10a + 15b + 20c = 1500

Relationship:

a − c = 20

Step 3: Eliminate a

Multiply the total quantity equation by 10:

10a + 10b + 10c = 1000

Subtract from the revenue equation:

5b + 10c = 500

Divide by 5:

b + 2c = 100

Step 4: Use the third equation

a = c + 20

Substitute into the total quantity equation:

c + 20 + b + c = 100

b + 2c = 80

Step 5: Interpret the contradiction

The system requires both:

b + 2c = 100

and:

b + 2c = 80

These cannot both be true.

Therefore, the sales records are inconsistent.

Answer: No solution exists for the stated information.

What could this mean for the business?

Possible explanations include incorrect sales totals, inaccurate revenue records, or an incorrect statement about the relationship between products A and C.

A system of equations can therefore help identify problems in business data, not just calculate unknown quantities.

A consistent version

Suppose the revenue was $1,400 rather than $1,500.

The revenue equation becomes:

10a + 15b + 20c = 1400

Subtracting 10 times the quantity equation gives:

b + 2c = 80

This is the same relationship obtained from a = c + 20.

The revised system has infinitely many solutions, not one unique solution.

An additional independent piece of information would be required to determine the exact quantities.

This illustrates an important limitation: three equations do not guarantee a unique solution if some equations contain redundant information.

Application: Production Planning with Three Products

A factory produces three types of furniture: chairs, tables, and shelves.

Each product requires a different amount of assembly time.

The factory manufactures 100 items using 290 assembly-hours.

It manufactures 10 more chairs than shelves.

Determine the number of each product.

Step 1: Define variables

c = chairs manufactured

t = tables manufactured

s = shelves manufactured

Step 2: Construct equations

Total items:

c + t + s = 100

Total assembly time:

2c + 4t + 3s = 290

Quantity relationship:

c − s = 10

Step 3: Eliminate c

Multiply the first equation by 2:

2c + 2t + 2s = 200

Subtract from the assembly-time equation:

2t + s = 90

Step 4: Use the relationship

c = s + 10

Substitute into the total quantity equation:

s + 10 + t + s = 100

t + 2s = 90

Step 5: Solve the reduced system

We have:

2t + s = 90

t + 2s = 90

Multiply the second equation by 2:

2t + 4s = 180

Subtract the first equation:

3s = 90

s = 30

Then:

t + 2(30) = 90

t = 30

And:

c = 30 + 10 = 40

Step 6: Verify

Total items:

40 + 30 + 30 = 100

Assembly time:

2(40) + 4(30) + 3(30)

= 80 + 120 + 90

= 290 hours

Quantity relationship:

40 − 30 = 10

Answer: 40 chairs, 30 tables, and 30 shelves.

Business conclusion

The factory's production records are consistent with manufacturing 40 chairs, 30 tables, and 30 shelves.

This information could be used to estimate material requirements, labour costs, and future production capacity.

Application: Profit Targets

A business sells handmade bags for $35 each.

The variable cost is $15 per bag.

Monthly fixed costs are $2,400.

The owner wants to earn a monthly profit of $1,600.

How many bags must be sold?

Step 1: Define the variable

x = number of bags sold

Step 2: Write the revenue equation

R = 35x

Step 3: Write the cost equation

C = 2400 + 15x

Step 4: Write the profit equation

P = R − C

P = 35x − (2400 + 15x)

P = 20x − 2400

Step 5: Set the target profit

1600 = 20x − 2400

4000 = 20x

x = 200

Step 6: Verify

Revenue:

35(200) = $7000

Total cost:

2400 + 15(200) = $5400

Profit:

7000 − 5400 = $1600

Answer: The business must sell 200 bags to achieve the target profit of $1,600.

Understanding Contribution Margin

The contribution margin per unit is the selling price minus the variable cost per unit.

Contribution margin = p − v

For this business:

35 − 15 = $20 per bag

Each bag sold contributes $20 toward covering fixed costs and generating profit.

This allows us to write:

P = (p − v)x − F

For a target profit T:

x = (F + T)/(p − v)

provided p − v is positive.

In this example:

x = (2400 + 1600)/(35 − 15)

x = 4000/20

x = 200 bags

When the answer is not a whole number

Suppose the calculation gives:

x = 152.4

If x represents complete products, the business generally needs to sell at least 153 products to reach or exceed its target, assuming the contribution margin is positive.

Rounding down would not achieve the required profit.

Application: Comparing Profit Between Two Businesses

Two businesses sell the same type of product.

Business A has:

• Selling price: $25 per unit

• Variable cost: $10 per unit

• Fixed cost: $1,500

Business B has:

• Selling price: $25 per unit

• Variable cost: $7 per unit

• Fixed cost: $2,400

At what sales volume do the businesses earn the same profit?

Step 1: Construct the profit equations

Business A:

Pₐ = 25x − (1500 + 10x)

Pₐ = 15x − 1500

Business B:

Pᵦ = 25x − (2400 + 7x)

Pᵦ = 18x − 2400

Step 2: Set the profits equal

15x − 1500 = 18x − 2400

900 = 3x

x = 300

Step 3: Calculate the profit

Business A:

Pₐ = 15(300) − 1500 = $3000

Business B:

Pᵦ = 18(300) − 2400 = $3000

Both businesses earn $3,000 profit at 300 units sold.

Step 4: Interpret

Below 300 units, Business A earns more profit because it has lower fixed costs.

Above 300 units, Business B earns more profit because its variable cost per unit is lower.

Comparison table

Business conclusion

The better business model depends on expected sales volume.

A business expecting relatively low sales may prefer lower fixed costs, while one expecting higher sales may benefit from lower variable costs.

Applications in Economics: Supply and Demand

Systems of equations are also used to model markets.

Two important economic concepts are supply and demand.

Demand

Demand describes how much of a product consumers are willing and able to purchase at different prices.

In a simple linear model, quantity demanded decreases as price increases.

For example:

Qd = 100 − 2p

where:

Qd = quantity demanded

p = price

Supply

Supply describes how much producers are willing and able to sell at different prices.

In a simple linear model, quantity supplied increases as price increases.

For example:

Qs = 20 + 2p

Market Equilibrium

Market equilibrium occurs when:

Qd = Qs

This is the price at which quantity demanded equals quantity supplied in the model.

Worked Example: Finding Equilibrium Price

Demand equation:

Qd = 100 − 2p

Supply equation:

Qs = 20 + 2p

At equilibrium:

100 − 2p = 20 + 2p

80 = 4p

p = 20

Substitute into demand:

Qd = 100 − 2(20)

Qd = 60

Substitute into supply:

Qs = 20 + 2(20)

Qs = 60

Equilibrium price = $20

Equilibrium quantity = 60 units

Interpreting the result

At a price of $20, consumers demand 60 units and producers supply 60 units.

At prices below equilibrium, the model predicts that quantity demanded exceeds quantity supplied.

At prices above equilibrium, the model predicts that quantity supplied exceeds quantity demanded.

Important limitation

Real markets are influenced by many factors, including consumer preferences, competition, income, taxes, and changes in production costs.

Linear supply-and-demand equations are simplified models, not exact predictions of every market.

Application: Comparing Subscription Plans

A business is choosing between two software subscriptions.

Plan A charges:

$50 per month plus $4 per user.

Plan B charges:

$110 per month plus $2 per user.

Find the number of users at which the plans cost the same.

Step 1: Define the variable

x = number of users

Step 2: Construct equations

Plan A:

Cₐ = 50 + 4x

Plan B:

Cᵦ = 110 + 2x

Step 3: Set costs equal

50 + 4x = 110 + 2x

2x = 60

x = 30

Step 4: Verify

Plan A:

50 + 4(30) = $170

Plan B:

110 + 2(30) = $170

Answer: Both plans cost $170 per month for 30 users.

Business interpretation

For fewer than 30 users, Plan A is cheaper.

For more than 30 users, Plan B is cheaper.

At exactly 30 users, both plans have the same monthly cost.

This type of comparison is useful when choosing subscriptions, equipment leases, service providers, or production arrangements.

Using Technology in Business Modelling

Spreadsheets and graphing software are useful for business analysis.

For example, a spreadsheet can calculate revenue, cost, and profit for many different sales volumes.

Consider:

R = 18x

C = 1500 + 8x

P = 10x − 1500

A spreadsheet could generate the following table.

The table makes the break-even point easy to identify.

Using graphing software

You can enter:

y = 18x

y = 1500 + 8x

into Desmos Graphing Calculator or GeoGebra Graphing Calculator.

The intersection represents the break-even point.

Technology is particularly useful when businesses need to compare several scenarios or change assumptions.

Evaluating the Reasonableness of Business Solutions

A mathematical answer must make sense in the business context.

Check 1: Units

If x represents the number of products sold, the answer should be expressed in units, not dollars.

If P represents profit, the answer should be expressed in money.

Check 2: Whole numbers

If a calculation gives:

x = 125.6 products

the business cannot normally sell 0.6 of a complete product.

If the goal is to meet or exceed a minimum profit, round up to 126 products when the model supports doing so.

Check 3: Negative values

A negative profit represents a loss.

However, a negative number of products sold is usually invalid in a simple sales model.

Check 4: Capacity

A company may calculate that it needs to sell 5,000 units to achieve its target.

But if it can only produce 3,000 units, the target cannot be reached under the current model.

Check 5: Model assumptions

A linear model assumes constant rates.

Real businesses may experience:

• Bulk discounts.

• Changes in labour costs.

• Limited production capacity.

• Seasonal demand.

• Changing selling prices.

• Additional fixed costs at higher output levels.

These factors can make a simple linear model less accurate.

Check 6: Verification

Always substitute calculated values into the original equations.

For example, if a system predicts that 40 units of A and 30 units of B generate $1,100:

20(40) + 10(30) = 1100

800 + 300 = 1100

The revenue equation is satisfied.

But other conditions must also be checked.

Communicating Mathematical Conclusions

In business and economics, a numerical answer is often not enough.

A complete conclusion should explain what the result means and how it can support a decision.

Example: Weak conclusion

x = 300

Improved conclusion

The two production plans have equal costs when 300 units are manufactured.

Strong conclusion

The two production plans each cost $3,500 at a production level of 300 units. Plan B is cheaper below 300 units because it has lower fixed costs, while Plan A is cheaper above 300 units because it has lower variable costs. Therefore, the company should consider its expected production volume before choosing a plan.

The final conclusion connects the mathematical result to a practical business decision.

A useful structure

When communicating a business solution, include:

1. Result: State the numerical answer with units.

2. Meaning: Explain what the value represents.

3. Verification: Show that it satisfies the original conditions.

4. Recommendation: Explain what the result suggests.

5. Limitation: Identify any assumptions that could affect the decision.

Common Mistakes

Confusing revenue with profit

Revenue is money received from sales. Profit is revenue minus total costs.

Forgetting fixed costs

A business may lose money even when each product sells for more than its variable production cost.

Using the wrong break-even equation

Break-even occurs when total revenue equals total cost, not when revenue equals variable cost alone.

Mixing prices and quantities

The selling price of a product and the number of products sold are different variables.

Constructing incorrect revenue equations

Revenue is calculated by multiplying price by quantity.

Forgetting to include all products

When several products are sold, total revenue must include the revenue from every product.

Rounding sales quantities incorrectly

When calculating the minimum number of complete units required to reach a profit target, round up when appropriate.

Assuming every system has a unique solution

Some business systems contain redundant or contradictory information.

Ignoring production capacity

A mathematically possible sales target may be impossible with existing resources.

Accepting unrealistic negative quantities

Negative profit may be meaningful, but negative quantities of physical products are generally not.

Failing to explain the result

Business conclusions should describe the meaning of the answer, not just provide a number.

Ignoring assumptions

Constant selling prices and costs may not remain realistic over large production ranges.

Key Terms

• Fixed cost: A cost that remains constant over the relevant production range.

• Variable cost: A cost that changes with production volume.

• Total cost: The combined fixed and variable costs.

• Revenue: Money received from sales.

• Profit: Revenue minus total cost.

• Loss: A negative profit.

• Break-even point: The sales volume at which revenue equals total cost.

• Contribution margin: Selling price minus variable cost per unit.

• Selling price: The amount charged for a product or service.

• Unit cost: The cost associated with one unit.

• Production volume: The number of units manufactured.

• Sales volume: The number of units sold.

• Profit target: A desired amount of profit.

• Cost model: An equation representing how costs depend on production or activity.

• Revenue model: An equation representing income from sales.

• Supply: The quantity producers are willing and able to sell at different prices.

• Demand: The quantity consumers are willing and able to buy at different prices.

• Market equilibrium: The point where quantity supplied equals quantity demanded.

• Constraint: A restriction on possible values or decisions.

• Mathematical model: A representation of a real-world situation using equations.

• Verification: Checking that a solution satisfies the original equations.

• Reasonableness: Evaluating whether a mathematical result makes sense in context.

Key Takeaways

• Systems of equations can model costs, revenue, profit, and production.

• Total cost can be represented by C = F + vx.

• Revenue can be represented by R = px when selling price is constant.

• Profit is calculated using P = R − C.

• Break-even occurs when revenue equals total cost.

• Systems can compare two business plans and identify when their costs or profits are equal.

• Two or more sales records can help determine unknown prices or quantities.

• Three-variable systems can model production involving several products.

• Supply-and-demand systems can determine market equilibrium in simplified economic models.

• Mathematical solutions should be interpreted using appropriate units.

• Business decisions should consider capacity, realistic quantities, and model assumptions.

• Strong conclusions explain what a solution means and how it supports a decision.

Check Your Understanding

Recall and Understanding

1. Explain the difference between fixed costs and variable costs. Give one example of each.

2. Define revenue, cost, and profit.

3. What does the break-even point represent?

4. Explain why a business might prefer a production method with higher fixed costs but lower variable costs.

5. Why should a business evaluate the assumptions of a mathematical model before using it to make decisions?

Cost and Profit Problems

6. A company sells water bottles for $20 each. Variable costs are $8 per bottle, and fixed costs are $1,200.

a. Write the revenue equation.

b. Write the total cost equation.

c. Write the profit equation.

d. Find the break-even quantity.

7. A business sells products for $30 each. Variable costs are $12 per unit, and fixed costs are $1,800.

How many products must be sold to earn a profit of $1,800?

8. A manufacturer has fixed costs of $750 and variable costs of $6 per unit.

Calculate the total cost of producing 200 units.

9. A company records total costs of $1,500 when producing 100 units and $2,700 when producing 300 units.

Assuming a linear cost model, determine the fixed cost and variable cost per unit.

10. A company sells 250 products at $24 each. Total costs are $4,200.

Calculate revenue and profit.

Systems and Business Applications

11. Two companies offer production services.

Plan A: $1,000 fixed cost plus $8 per unit.

Plan B: $400 fixed cost plus $12 per unit.

Find the production quantity at which the plans cost the same.

12. A school sells adult tickets for $15 and student tickets for $10.

A total of 120 tickets are sold, generating $1,500.

Determine the number of each ticket type.

13. A shop sells notebooks and pens.

Three notebooks and two pens cost $23.

Two notebooks and four pens cost $26.

Find the price of each item.

14. A factory manufactures chairs, tables, and shelves.

It produces 100 items using 290 assembly-hours.

Each chair requires 2 hours, each table 4 hours, and each shelf 3 hours.

The factory produces 10 more chairs than shelves.

Determine the number of each product.

15. The demand and supply equations for a product are:

Qd = 120 − 3p

Qs = 20 + 2p

Find the equilibrium price and quantity.

Challenge

16. A company sells a product for $25 per unit. Its fixed costs are $2,000, and its variable costs are $10 per unit.

The company can produce at most 180 units.

Can it achieve a profit of $1,000? Explain using calculations.

17. A business compares two profit models:

Pₐ = 12x − 900

Pᵦ = 15x − 1500

Determine when the profits are equal and which model is better for higher sales volumes.

18. A company sells two products.

Product A sells for $20 and Product B for $30.

The company sells 100 units and earns $2,400.

Find the quantity of each product.

19. A manufacturer calculates that it must sell 175.4 units to reach a profit target.

Explain how the answer should be interpreted if only complete units can be sold.

20. A business reports:

x + y + z = 100

2x + 3y + 4z = 290

x − z = 10

Solve the system and explain what the solution could represent in a production context.

Answers to Check Your Understanding

1. Fixed and variable costs

Fixed costs remain unchanged over the relevant production range, such as monthly rent.

Variable costs change with the number of units produced, such as raw materials or packaging.

2. Revenue, cost, and profit

Revenue is money received from sales.

Cost is money spent to operate the business or produce goods.

Profit is the difference:

P = R − C

3. Break-even point

The break-even point is where total revenue equals total cost.

At this point, profit is zero.

4. Higher fixed costs and lower variable costs

A company with higher fixed costs but lower variable costs may become cheaper at high production volumes.

The lower variable cost can eventually offset the higher fixed cost.

5. Evaluating assumptions

Mathematical models simplify real situations.

For example, a linear model may assume constant selling prices, fixed costs, and variable costs.

If these assumptions do not hold, the model's predictions may become inaccurate.

6. Water bottle business

Given:

Selling price = $20

Variable cost = $8

Fixed cost = $1,200

a. Revenue:

R = 20x

b. Total cost:

C = 1200 + 8x

c. Profit:

P = 20x − (1200 + 8x)

P = 12x − 1200

d. Break-even:

20x = 1200 + 8x

12x = 1200

x = 100

Answer: The business breaks even at 100 bottles.

Verification:

Revenue = 20(100) = $2000

Cost = 1200 + 8(100) = $2000

7. Profit target

Selling price = $30

Variable cost = $12

Fixed cost = $1800

Target profit = $1800

Profit equation:

P = 30x − (1800 + 12x)

P = 18x − 1800

Set the target:

1800 = 18x − 1800

3600 = 18x

x = 200

Answer: 200 products.

Verification:

Revenue = 30(200) = $6000

Cost = 1800 + 12(200) = $4200

Profit = $1800

8. Total production cost

C = F + vx

C = 750 + 6(200)

C = 750 + 1200

Answer: $1,950.

9. Finding fixed and variable costs

Let:

F = fixed cost

v = variable cost per unit

Construct equations:

F + 100v = 1500

F + 300v = 2700

Subtract:

200v = 1200

v = 6

Substitute:

F + 100(6) = 1500

F = 900

Answer: Fixed cost = $900; variable cost = $6 per unit.

Cost equation:

C = 900 + 6x

10. Revenue and profit

Revenue:

R = 250(24)

R = $6000

Profit:

P = R − C

P = 6000 − 4200

Answer: Revenue = $6,000; profit = $1,800.

11. Comparing production plans

Plan A:

Cₐ = 1000 + 8x

Plan B:

Cᵦ = 400 + 12x

Set costs equal:

1000 + 8x = 400 + 12x

600 = 4x

x = 150

Calculate cost:

1000 + 8(150) = $2200

400 + 12(150) = $2200

Answer: Both plans cost $2,200 at 150 units.

Below 150 units, Plan B is cheaper.

Above 150 units, Plan A is cheaper.

12. Ticket sales

Let:

a = adult tickets

s = student tickets

Equations:

a + s = 120

15a + 10s = 1500

Multiply the first equation by 10:

10a + 10s = 1200

Subtract:

5a = 300

a = 60

Then:

s = 120 − 60

s = 60

Answer: 60 adult tickets and 60 student tickets.

Verification:

60 + 60 = 120

15(60) + 10(60) = $1500

13. Notebook and pen prices

Let:

n = notebook price

p = pen price

Equations:

3n + 2p = 23

2n + 4p = 26

Multiply the first equation by 2:

6n + 4p = 46

Subtract the second equation:

4n = 20

n = 5

Substitute:

3(5) + 2p = 23

2p = 8

p = 4

Answer: Notebook = $5; pen = $4.

Verification:

3(5) + 2(4) = 23

2(5) + 4(4) = 26

14. Furniture production

Let:

c = chairs

t = tables

s = shelves

Equations:

c + t + s = 100

2c + 4t + 3s = 290

c − s = 10

From the third equation:

c = s + 10

Substitute into the total quantity equation:

t + 2s = 90

Substitute into the assembly-time equation:

2(s + 10) + 4t + 3s = 290

4t + 5s = 270

Multiply t + 2s = 90 by 4:

4t + 8s = 360

Subtract:

3s = 90

s = 30

Then:

t = 90 − 2(30) = 30

c = 30 + 10 = 40

Answer: 40 chairs, 30 tables, and 30 shelves.

Verification:

40 + 30 + 30 = 100

2(40) + 4(30) + 3(30) = 290

15. Market equilibrium

Demand:

Qd = 120 − 3p

Supply:

Qs = 20 + 2p

At equilibrium:

120 − 3p = 20 + 2p

100 = 5p

p = 20

Substitute:

Q = 120 − 3(20)

Q = 60

Answer: Equilibrium price = $20; equilibrium quantity = 60 units.

Verification:

Demand = 60

Supply = 20 + 2(20) = 60

16. Profit target and capacity

Selling price = $25

Variable cost = $10

Fixed cost = $2000

Maximum production = 180 units

Profit equation:

P = 25x − (2000 + 10x)

P = 15x − 2000

To earn $1000:

1000 = 15x − 2000

3000 = 15x

x = 200

However, the company can only produce 180 units.

Maximum profit at full capacity:

P = 15(180) − 2000

P = 2700 − 2000

P = $700

Answer: No. The company cannot achieve a $1,000 profit under the stated conditions. Its maximum possible profit is $700 if it produces and sells all 180 units.

17. Comparing profit models

Pₐ = 12x − 900

Pᵦ = 15x − 1500

Set equal:

12x − 900 = 15x − 1500

600 = 3x

x = 200

Calculate profit:

Pₐ = 12(200) − 900 = $1500

Pᵦ = 15(200) − 1500 = $1500

Answer: Both models produce $1,500 profit at 200 units.

For more than 200 units, Model B produces greater profit because its profit increases by $15 per additional unit compared with $12 for Model A.

18. Two-product sales

Let:

a = units of Product A

b = units of Product B

Total quantity:

a + b = 100

Revenue:

20a + 30b = 2400

Multiply the first equation by 20:

20a + 20b = 2000

Subtract:

10b = 400

b = 40

Then:

a = 60

Answer: 60 units of Product A and 40 units of Product B.

Verification:

60 + 40 = 100

20(60) + 30(40) = $2400

19. Fractional sales quantity

The calculated quantity is:

x = 175.4

If the business must sell complete units and needs to reach or exceed a profit target, it should round up.

Answer: The business must sell at least 176 units, assuming profit increases with each additional unit and the model remains valid.

Rounding down to 175 would leave the business below its target.

20. Three-product production

Given:

x + y + z = 100

2x + 3y + 4z = 290

x − z = 10

From the third equation:

x = z + 10

Substitute into the first:

(z + 10) + y + z = 100

y + 2z = 90

Substitute into the second:

2(z + 10) + 3y + 4z = 290

3y + 6z = 270

Divide by 3:

y + 2z = 90

Both reduced equations are identical.

Therefore, the system does not have a unique solution.

Let z = t.

Then:

x = t + 10

y = 90 − 2t

The general solution is:

(x, y, z) = (t + 10, 90 − 2t, t)

For nonnegative quantities:

0 ≤ t ≤ 45

If the variables represent complete products, t must be an integer.

For example, one possible production combination is:

x = 40

y = 30

z = 30

Verification:

40 + 30 + 30 = 100

2(40) + 3(30) + 4(30) = 290

40 − 30 = 10

Answer: The system has infinitely many nonnegative real solutions and 46 possible whole-number production combinations.

An additional independent condition would be needed to determine the exact production quantities.

Final Summary

Systems of equations provide a practical foundation for analysing business and economic situations.

They can be used to calculate costs, determine prices, establish profit targets, identify break-even points, compare business strategies, and investigate supply-and-demand relationships.

However, mathematical calculations are only part of effective business decision-making. Results must be verified, interpreted using appropriate units, and evaluated against real-world constraints.

The most valuable business solution is not simply a correct number, but a well-supported conclusion that helps explain a situation and guide a decision.

2. Science Applications

Introduction

Science uses mathematics to describe, explain, and predict how the natural world behaves. Whether scientists are studying the motion of objects, analysing chemical reactions, measuring electrical circuits, or investigating experimental data, they frequently encounter situations involving several unknown quantities.

A system of equations allows scientists to combine different pieces of information and determine unknown values.

For example, a scientist might need to determine:

• The masses of two substances in a mixture.

• The speeds of two moving objects.

• The forces acting on an object in equilibrium.

• The currents flowing through different parts of an electrical circuit.

• The concentrations of solutions prepared in a laboratory.

• The unknown parameters in an experimental model.

Each measurement or scientific law provides a mathematical relationship. When enough independent relationships are available, the unknown quantities can often be determined.

Systems of equations are especially useful because they allow scientists to analyse several connected quantities simultaneously.

However, scientific models are simplified representations of reality. A mathematically correct solution must still be evaluated using scientific principles, appropriate units, experimental uncertainty, and the assumptions of the model.

Why Are Systems of Equations Important in Science?

Scientific investigations often involve multiple unknown quantities.

Consider a laboratory mixture containing two liquids.

A scientist knows:

• The total volume of the mixture.

• The total mass of the mixture.

• The densities of the two liquids.

The scientist wants to determine the volume of each liquid.

One equation can describe the total volume.

Another equation can describe the total mass.

Together, these equations form a system that can determine both unknown volumes.

This approach is used across many scientific disciplines.

The general process is similar in every field:

Scientific information → Mathematical equations → Solution → Scientific interpretation → Verification

Building a Mathematical Model from Scientific Information

A mathematical model represents a scientific situation using variables and equations.

A reliable modelling process involves several steps.

Step 1: Identify the unknown quantities

Define variables clearly and include their units.

For example:

m₁ = mass of substance A, in grams

m₂ = mass of substance B, in grams

Step 2: Identify the scientific relationships

Look for relevant scientific laws or definitions.

Examples include:

Speed = distance/time

Density = mass/volume

Force = mass × acceleration

Charge = current × time

Concentration = mass of solute/volume of solution

Step 3: Construct equations

Use the information in the problem to create relationships between the variables.

Step 4: Solve the system

Choose an appropriate method:

• Substitution.

• Elimination.

• Graphing.

• Matrix methods.

• Technology.

Step 5: Interpret the solution

Explain what each value represents and include correct units.

Step 6: Verify the result

Substitute the values into the original equations and check whether they satisfy the scientific conditions.

Step 7: Evaluate the model

Consider whether the assumptions are realistic and whether the measurements are sufficiently accurate.

Application: Mixing Two Liquids Using Density

Density describes how much mass is contained in a given volume.

The density equation is:

ρ = m/V

where:

• ρ = density.

• m = mass.

• V = volume.

Rearranging:

m = ρV

Suppose a laboratory technician mixes two liquids.

Liquid A has a density of 0.8 g/mL.

Liquid B has a density of 1.2 g/mL.

The mixture has:

• Total volume = 100 mL.

• Total mass = 104 g.

Assuming the volumes are additive, determine the volume of each liquid.

Step 1: Define variables

x = volume of Liquid A, in mL

y = volume of Liquid B, in mL

Step 2: Construct the volume equation

The total volume is 100 mL.

x + y = 100

Step 3: Construct the mass equation

Mass = density × volume

Liquid A has mass:

0.8x

Liquid B has mass:

1.2y

The total mass is 104 g.

0.8x + 1.2y = 104

Step 4: Solve by substitution

From the volume equation:

x = 100 − y

Substitute:

0.8(100 − y) + 1.2y = 104

80 − 0.8y + 1.2y = 104

0.4y = 24

y = 60

Then:

x = 100 − 60

x = 40

Step 5: Interpret

Liquid A: 40 mL

Liquid B: 60 mL

Step 6: Verify

Total volume:

40 + 60 = 100 mL

Total mass:

0.8(40) + 1.2(60)

= 32 + 72

= 104 g

Both conditions are satisfied.

Answer: The mixture contains 40 mL of Liquid A and 60 mL of Liquid B.

Limitations of the model

The calculation assumes that the volumes are additive.

In real mixtures, molecular interactions can cause the final volume to differ from the sum of the separate volumes.

The model also assumes that the densities remain constant at the experimental temperature.

These assumptions must be considered before applying the result to an actual laboratory mixture.

Application: Preparing Solutions of Different Concentrations

Scientists often prepare solutions by combining liquids with different concentrations.

Suppose a laboratory needs 500 mL of a 20% solution.

Two solutions are available:

• Solution A: 10%.

• Solution B: 30%.

How much of each solution is required?

Step 1: Define variables

x = volume of 10% solution, in mL

y = volume of 30% solution, in mL

Step 2: Total volume equation

x + y = 500

Step 3: Solute equation

The amount of solute in each solution is proportional to its concentration.

0.10x + 0.30y = 0.20(500)

Therefore:

0.10x + 0.30y = 100

The equation uses equivalent volumes of solute under a simplified volume-percent model.

Step 4: Solve

From the first equation:

x = 500 − y

Substitute:

0.10(500 − y) + 0.30y = 100

50 − 0.10y + 0.30y = 100

0.20y = 50

y = 250

Then:

x = 500 − 250

x = 250

Step 5: Interpret

250 mL of the 10% solution

250 mL of the 30% solution

Step 6: Verify

Total volume:

250 + 250 = 500 mL

Solute-equivalent volume:

0.10(250) + 0.30(250)

= 25 + 75

= 100 mL

Final concentration:

100/500 × 100% = 20%

Answer: Mix 250 mL of each solution.

Scientific interpretation

The required concentration lies halfway between 10% and 30%.

Therefore, equal volumes produce the desired 20% concentration under the model's assumptions.

Important limitation

This simplified model assumes consistent concentration definitions and additive volumes. In laboratory work, concentration units such as mol/L, g/L, and mass percent must not be mixed without conversion.

For precise solution preparation, a laboratory may need to measure solute quantity and dilute to a specified final volume rather than simply add two measured volumes.

Application: Motion of Two Objects

Systems of equations can be used to analyse objects moving at constant speeds.

The basic relationship is:

d = vt

where:

• d = distance.

• v = speed.

• t = time.

Worked Example: Two Cyclists

Two cyclists begin travelling from the same location in opposite directions.

After 3 hours, they are 90 km apart.

One cyclist travels 6 km/h faster than the other.

Find their speeds.

Step 1: Define variables

x = speed of the slower cyclist, in km/h

y = speed of the faster cyclist, in km/h

Step 2: Construct the distance equation

The slower cyclist travels:

3x kilometres

The faster cyclist travels:

3y kilometres

Since they travel in opposite directions, their separation is the sum of the distances.

3x + 3y = 90

Divide by 3:

x + y = 30

Step 3: Construct the speed equation

The faster cyclist travels 6 km/h faster.

y − x = 6

Step 4: Solve

Add the equations:

x + y = 30

−x + y = 6

2y = 36

y = 18

Then:

x = 30 − 18

x = 12

Step 5: Verify

Slower cyclist:

d = 12(3) = 36 km

Faster cyclist:

d = 18(3) = 54 km

Total separation:

36 + 54 = 90 km

Speed difference:

18 − 12 = 6 km/h

Answer: The slower cyclist travels at 12 km/h, and the faster cyclist travels at 18 km/h.

Limitations of the model

The calculation assumes that both cyclists:

• Travel in exactly opposite directions along the same straight route.

• Maintain constant speeds.

• Start at the same time.

• Do not stop during the journey.

Real journeys may involve acceleration, changes in direction, and stops.

The model is therefore an approximation of actual motion.

Application: Forces in Equilibrium

An object is in equilibrium when the resultant force acting on it is zero.

For equilibrium in two dimensions:

ΣFₓ = 0

ΣFᵧ = 0

These equations can be used together to determine unknown forces.

Worked Example: A Hanging Sign

A sign weighing 100 N is supported by two cables.

The left cable makes an angle of 30° above the horizontal.

The right cable makes an angle of 60° above the horizontal.

Determine the tension in each cable.

Step 1: Define variables

T₁ = tension in the left cable, in N

T₂ = tension in the right cable, in N

Step 2: Resolve the forces horizontally

The horizontal components must balance.

The left cable pulls left.

The right cable pulls right.

T₁ cos 30° = T₂ cos 60°

Using:

cos 30° ≈ 0.866

cos 60° = 0.5

We obtain:

0.866T₁ = 0.5T₂

Therefore:

T₂ ≈ 1.732T₁

Step 3: Resolve the forces vertically

The upward components must balance the 100 N weight.

T₁ sin 30° + T₂ sin 60° = 100

Using:

sin 30° = 0.5

sin 60° ≈ 0.866

We obtain:

0.5T₁ + 0.866T₂ = 100

Step 4: Substitute

0.5T₁ + 0.866(1.732T₁) ≈ 100

0.5T₁ + 1.5T₁ ≈ 100

2T₁ ≈ 100

T₁ ≈ 50 N

Then:

T₂ ≈ 1.732(50)

T₂ ≈ 86.6 N

Step 5: Interpret

Left cable tension ≈ 50 N

Right cable tension ≈ 86.6 N

The right cable has greater tension because its steeper angle changes how the force is distributed between horizontal and vertical components.

Step 6: Verify

Horizontal:

50 cos 30° ≈ 86.6 cos 60°

43.3 N ≈ 43.3 N

Vertical:

50 sin 30° + 86.6 sin 60°

≈ 25 + 75

≈ 100 N

Answer: T₁ = 50 N and T₂ ≈ 86.6 N.

Scientific assumptions

This model assumes:

• The sign is stationary.

• The cables are massless.

• The cable angles are known accurately.

• No additional horizontal forces act.

• The sign is treated as a particle at the support junction, or the support geometry also satisfies rotational equilibrium.

In real engineering problems, the sign's dimensions, attachment points, and torque balance may also need to be considered.

Application: Electrical Circuits

Electrical circuits can be analysed using systems of equations.

Two important principles are:

Kirchhoff's Current Law

The total current entering a junction equals the total current leaving the junction.

Kirchhoff's Voltage Law

The sum of voltage changes around a closed circuit loop equals zero.

Ohm's Law is also important:

V = IR

where:

• V = potential difference, in volts.

• I = current, in amperes.

• R = resistance, in ohms.

Worked Example: Currents in a Circuit

At a junction, a total current of 6 A splits into two branches.

The first branch has resistance 4 Ω.

The second branch has resistance 8 Ω.

Both branches are connected in parallel, so the potential difference across each branch is equal.

Find the current in each branch.

Step 1: Define variables

I₁ = current through the 4 Ω branch, in A

I₂ = current through the 8 Ω branch, in A

Step 2: Apply the current law

I₁ + I₂ = 6

Step 3: Apply Ohm's Law

The voltage across each branch is the same.

4I₁ = 8I₂

Divide by 4:

I₁ = 2I₂

Step 4: Solve

Substitute:

2I₂ + I₂ = 6

3I₂ = 6

I₂ = 2 A

Then:

I₁ = 4 A

Step 5: Verify

Total current:

4 + 2 = 6 A

Voltage across first branch:

V₁ = 4(4) = 16 V

Voltage across second branch:

V₂ = 8(2) = 16 V

Both branches have the same potential difference.

Answer: The 4 Ω branch carries 4 A, and the 8 Ω branch carries 2 A.

Scientific interpretation

The branch with lower resistance carries more current when the voltage is the same.

This agrees with Ohm's Law.

Model limitations

The calculation assumes ideal resistors with constant resistance.

In real circuits, resistance can change with temperature, and components may not behave ideally.

Application: Determining Unknown Quantities from Experimental Data

Scientific experiments often produce measurements that can be used to construct equations.

Consider a spring that follows Hooke's Law:

F = kx

where:

• F = applied force, in newtons.

• k = spring constant, in N/m.

• x = extension, in metres.

A real experimental setup may also contain a constant force offset caused by calibration error or another systematic effect.

A more general linear model is:

F = kx + b

where b represents the force offset.

Worked Example: Finding the Spring Constant

An experiment produces the following data:

Assume the relationship is linear.

Find k and b.

Step 1: Define variables

k = spring constant, in N/m

b = force offset, in N

Step 2: Construct equations

First measurement:

3 = 0.02k + b

Second measurement:

7 = 0.06k + b

Step 3: Eliminate b

Subtract:

7 − 3 = (0.06 − 0.02)k

4 = 0.04k

k = 100 N/m

Step 4: Find b

3 = 0.02(100) + b

3 = 2 + b

b = 1 N

Step 5: Write the model

F = 100x + 1

where F is in N and x is in m.

Step 6: Interpret

The gradient is 100 N/m.

This represents the spring constant in the linear model.

The intercept is 1 N.

This indicates that the measured force includes a constant offset.

Step 7: Verify

At x = 0.02 m:

F = 100(0.02) + 1 = 3 N

At x = 0.06 m:

F = 100(0.06) + 1 = 7 N

Both measurements are reproduced.

Limitations

Two data points always determine a straight line when their x-values are different.

However, this does not prove that the physical relationship is perfectly linear.

Scientists should collect additional measurements and examine whether the model remains appropriate.

The spring may also stop following Hooke's Law beyond its elastic limit.

Application: Determining a Linear Motion Model

A moving object has position described by:

s = vt + s₀

where:

• s = position, in metres.

• v = constant velocity, in m/s.

• t = time, in seconds.

• s₀ = initial position, in metres.

A motion sensor records:

At t = 2 s, s = 11 m.

At t = 5 s, s = 23 m.

Determine the velocity and initial position.

Step 1: Construct equations

11 = 2v + s₀

23 = 5v + s₀

Step 2: Eliminate s₀

Subtract:

12 = 3v

v = 4 m/s

Step 3: Find initial position

11 = 2(4) + s₀

s₀ = 3 m

Step 4: Write the motion model

s = 4t + 3

Step 5: Verify

At t = 2 s:

s = 4(2) + 3 = 11 m

At t = 5 s:

s = 4(5) + 3 = 23 m

Answer: Velocity = 4 m/s; initial position = 3 m.

Scientific interpretation

The object moves in the positive direction at a constant velocity of 4 m/s.

At t = 0, its position is 3 m.

Limitation

The model assumes constant velocity.

If the object accelerates, a linear position-time equation may no longer be appropriate.

Application: Biology and Population Studies

Systems of equations can be used to estimate populations when several measurements are available.

Worked Example: Two Species in a Habitat

An ecological survey identifies two species of animals in a protected habitat.

Species A has 4 legs per animal.

Species B has 2 legs per animal.

The survey records:

• 50 animals in total.

• 160 legs in total.

Determine the number of animals of each species.

Step 1: Define variables

x = number of four-legged animals

y = number of two-legged animals

Step 2: Construct equations

Total animals:

x + y = 50

Total legs:

4x + 2y = 160

Step 3: Solve

Multiply the first equation by 2:

2x + 2y = 100

Subtract from the second:

2x = 60

x = 30

Then:

y = 50 − 30

y = 20

Step 4: Verify

Total animals:

30 + 20 = 50

Total legs:

4(30) + 2(20)

= 120 + 40

= 160

Answer: There are 30 four-legged animals and 20 two-legged animals.

Scientific limitations

The model assumes:

• Only the two specified species are present.

• Every animal has the expected number of legs.

• All animals are counted correctly.

• The recorded observations are accurate.

In a real ecological investigation, incomplete observations and sampling errors may affect the results.

Application: Chemical Reactions and Conservation of Mass

Chemical reactions obey the law of conservation of mass.

In a closed system, the total mass of reactants equals the total mass of products.

Systems of equations can be used to determine unknown masses.

Worked Example: Formation of a Compound

Two elements, A and B, react to form a compound.

The total mass of the compound is 60 g.

The mass of element A is twice the mass of element B.

Find the mass of each element.

Step 1: Define variables

x = mass of element A, in g

y = mass of element B, in g

Step 2: Construct equations

Conservation of mass:

x + y = 60

Mass relationship:

x = 2y

Step 3: Solve

Substitute:

2y + y = 60

3y = 60

y = 20

Then:

x = 40

Step 4: Verify

40 + 20 = 60 g

40 = 2(20)

Answer: Element A contributes 40 g and element B contributes 20 g.

Scientific interpretation

The compound contains 40 g of element A and 20 g of element B.

The mass ratio is:

A : B = 2 : 1

This example assumes complete reaction, no loss of material, and that the compound contains only the two elements.

In actual chemistry, chemical formulae and mole ratios must also be considered when determining how elements combine.

Application: Three-Variable Scientific Systems

More complex scientific situations may involve three unknown quantities.

Worked Example: Three Materials in a Mixture

A laboratory combines three materials, A, B, and C.

Their densities are:

The total volume is 100 mL.

The total mass is 210 g.

The volume of Material C is 10 mL greater than the volume of Material A.

Determine the volume of each material.

Step 1: Define variables

x = volume of Material A, in mL

y = volume of Material B, in mL

z = volume of Material C, in mL

Step 2: Construct equations

Total volume:

x + y + z = 100

Total mass:

x + 2y + 3z = 210

Volume relationship:

z − x = 10

Step 3: Eliminate x

Subtract the total volume equation from the mass equation:

y + 2z = 110

Step 4: Use the third equation

z = x + 10

Therefore:

x = z − 10

Substitute into the volume equation:

(z − 10) + y + z = 100

y + 2z = 110

The reduced equations are identical.

Step 5: Interpret

The system does not have a unique solution.

The third relationship is already implied by the total mass and volume equations.

We can express the solutions using z = t:

x = t − 10

y = 110 − 2t

z = t

For nonnegative volumes:

10 ≤ t ≤ 55

There are infinitely many nonnegative real solutions.

Scientific conclusion

The available measurements are insufficient to determine the exact volume of each material.

An additional independent measurement is required.

For example, measuring the volume of one component would allow the remaining volumes to be determined.

Why this example matters

Scientific investigations do not always provide enough information to determine every unknown.

Recognising insufficient information is an important part of scientific reasoning.

Improving Experimental Models with Additional Measurements

Suppose the scientist measures the volume of Material B and finds:

y = 30 mL

Using:

y + 2z = 110

30 + 2z = 110

2z = 80

z = 40 mL

Then:

x = z − 10

x = 30 mL

The volumes are:

Material A: 30 mL

Material B: 30 mL

Material C: 40 mL

Verify

Total volume:

30 + 30 + 40 = 100 mL

Total mass:

1(30) + 2(30) + 3(40)

= 30 + 60 + 120

= 210 g

Volume difference:

40 − 30 = 10 mL

Answer: 30 mL of A, 30 mL of B, and 40 mL of C.

The additional independent measurement allows a unique solution to be determined.

Using Experimental Data to Construct Systems

Scientists frequently collect measurements and use them to determine unknown parameters.

Consider a relationship of the form:

y = mx + b

where:

m = gradient

b = y-intercept

Suppose an experiment produces:

The equations are:

7 = 2m + b

16 = 5m + b

Subtract:

9 = 3m

m = 3

Then:

7 = 2(3) + b

b = 1

The model is:

y = 3x + 1

Interpreting the parameters

The gradient represents the rate of change of y with respect to x.

The intercept represents the predicted value of y when x = 0.

The scientific meaning and units of these parameters depend on what x and y represent.

For example, if y is distance in metres and x is time in seconds:

m = 3 m/s

b = 1 m

The equation becomes:

s = 3t + 1

Why More Experimental Data Is Often Better

Two measurements can determine the gradient and intercept of a straight line.

However, real experimental measurements contain uncertainty.

For example:

These points are close to the relationship:

s = 3t + 1

But they do not lie exactly on one straight line.

This is common in experimental science.

Rather than constructing a system using only two measurements, scientists often use a line of best fit.

A best-fit line considers all available measurements and reduces the influence of random experimental variation.

Least-squares modelling

For a linear model:

y = mx + b

the least-squares method chooses m and b to minimise the sum of squared differences between measured and predicted values.

These differences are called residuals.

The resulting calculation can be expressed as a system of equations for m and b.

For n data points (xᵢ, yᵢ), the least-squares normal equations are:

mΣxᵢ² + bΣxᵢ = Σxᵢyᵢ

mΣxᵢ + nb = Σyᵢ

Solving this two-variable system gives the least-squares gradient and intercept, provided the x-values are not all identical.

This is an important connection between systems of equations and scientific data analysis.

At an introductory level, graphing software or spreadsheets can perform the calculation without requiring students to derive the equations.

Evaluating the Limitations of Mathematical Models

A mathematical model is useful only when its assumptions are appropriate for the situation being studied.

Scientific models simplify reality so that relationships can be represented mathematically.

However, simplification introduces limitations.

Limitation 1: Measurement Uncertainty

Scientific measurements are not perfectly exact.

For example, a measuring cylinder might record a volume of:

50.0 ± 0.5 mL

The actual volume may differ slightly from the recorded value.

When measured quantities are used in a system of equations, the calculated solution also contains uncertainty.

Limitation 2: Simplified Physical Conditions

A motion model may assume constant velocity.

A force model may ignore air resistance.

A circuit model may assume constant resistance.

A mixture model may assume additive volumes.

These assumptions can make the equations easier to solve but may reduce accuracy.

Limitation 3: Limited Experimental Data

A system may contain insufficient independent information.

For example:

x + y = 10

2x + 2y = 20

The second equation provides no new information.

The system has infinitely many solutions.

Additional independent measurements are needed to determine unique values.

Limitation 4: Inconsistent Data

Experimental measurements may contradict one another.

For example:

x + y = 10

x + y = 12

These equations cannot both be satisfied exactly.

This may indicate measurement error, an incorrect model, or incompatible experimental conditions.

Limitation 5: Sensitivity to Measurement Errors

Some systems are particularly sensitive to small changes in measured values.

Consider:

x + y = 10

x + 1.001y = 10.002

Subtract:

0.001y = 0.002

y = 2

Then:

x = 8

Now suppose the second measured constant changes slightly:

x + 1.001y = 10.003

Subtract:

0.001y = 0.003

y = 3

Then:

x = 7

A change of only 0.001 in one measured value changes the solution from (8, 2) to (7, 3).

This occurs because the equations represent lines with very similar slopes.

Such systems are described as ill-conditioned.

They can be difficult to solve accurately when experimental measurements contain uncertainty.

Limitation 6: Real-World Constraints

Scientific quantities may have restrictions.

Examples include:

• Mass cannot be negative.

• Volume cannot be negative.

• Population counts must be whole numbers.

• Concentration cannot exceed certain physical limits.

• A spring cannot stretch indefinitely while continuing to obey Hooke's Law.

A mathematical solution that violates these restrictions may be physically invalid.

Comparing Mathematical and Scientific Validity

A mathematically correct answer satisfies the equations.

A scientifically valid answer must also be consistent with physical laws, units, assumptions, and experimental conditions.

This distinction is essential when using mathematics in scientific investigations.

Communicating Scientific Conclusions

A strong scientific conclusion should do more than state a numerical answer.

It should include:

1. The calculated result.

2. Appropriate units.

3. An explanation of what the result means.

4. Verification using the original information.

5. Relevant assumptions or limitations.

Example: Basic conclusion

x = 40

y = 60

Improved scientific conclusion

The mixture contains 40 mL of Liquid A and 60 mL of Liquid B.

Strong scientific conclusion

The calculated mixture contains 40 mL of Liquid A and 60 mL of Liquid B. These volumes total 100 mL and produce a combined mass of 104 g using the given densities. The result is valid under the assumptions that the volumes are additive and the densities remain constant at the experimental temperature.

The final version demonstrates mathematical accuracy and scientific understanding.

Common Mistakes

Defining variables without units

Always state whether a variable represents mass, volume, time, speed, force, or another quantity.

Using inconsistent units

Convert measurements before constructing equations.

For example, 500 mL = 0.5 L.

Confusing mass and weight

Mass is measured in kilograms, while weight is a force measured in newtons.

Constructing equations that do not match the scientific situation

Use the correct scientific law and carefully interpret the information.

Ignoring direction

Velocity, acceleration, and force may require positive or negative signs depending on the chosen coordinate system.

Assuming all equations are independent

Redundant equations do not provide additional information.

Rounding too early

Keep sufficient precision during calculations.

Accepting impossible values

Check whether the result satisfies physical restrictions.

Ignoring experimental uncertainty

Measured values are not perfectly exact.

Assuming a model applies under all conditions

A linear model may only be appropriate over a limited range.

Failing to verify results

Substitute the calculated values into every original equation.

Providing answers without scientific interpretation

A complete answer should explain what the values mean in context.

Key Terms

• Mathematical model: A representation of a scientific situation using mathematical relationships.

• System of equations: Two or more equations solved simultaneously.

• Variable: A symbol representing an unknown quantity.

• Substitution: Replacing a variable with an equivalent expression or known value.

• Elimination: Combining equations to remove a variable.

• Density: Mass per unit volume.

• Concentration: Amount of solute relative to the amount of solution.

• Velocity: Rate of change of displacement with time.

• Equilibrium: A state in which the resultant force is zero, with rotational equilibrium also required for an extended rigid body.

• Resultant force: The vector sum of all forces acting on an object.

• Tension: A pulling force transmitted through a cable, rope, or similar object.

• Ohm's Law: The relationship V = IR for an ohmic conductor.

• Conservation of mass: The principle that mass is conserved in a closed system.

• Gradient: The rate of change represented by the slope of a graph.

• Intercept: The point where a graph crosses an axis.

• Experimental uncertainty: The range or degree of doubt associated with a measurement.

• Residual: The difference between an observed value and a model's predicted value.

• Line of best fit: A line used to represent the overall trend in experimental data.

• Least-squares method: A method that minimises the sum of squared residuals.

• Independent equations: Equations that provide distinct mathematical information.

• Inconsistent system: A system with no solution.

• Ill-conditioned system: A system whose solution is highly sensitive to small changes in the input data.

• Verification: Checking whether a solution satisfies the original equations.

• Scientific validity: The extent to which a result is consistent with scientific principles and conditions.

Key Takeaways

• Systems of equations are widely used in scientific investigations.

• Scientific laws provide relationships that can be written as equations.

• Density and concentration problems often involve conservation of mass or solute.

• Motion problems can use relationships between distance, speed, and time.

• Force equilibrium requires balancing force components.

• Electrical circuits can be analysed using current, voltage, and resistance relationships.

• Experimental data can be used to determine unknown model parameters.

• Three-variable systems may require additional independent measurements.

• Mathematical solutions must be interpreted using appropriate scientific units.

• Experimental uncertainty and model assumptions affect the reliability of results.

• A mathematically correct solution is not necessarily scientifically valid.

• Scientific conclusions should include calculations, interpretation, verification, and limitations.

Check Your Understanding

Recall and Understanding

1. Explain why systems of equations are useful in scientific investigations.

2. What is the difference between a mathematically correct solution and a scientifically valid solution?

3. Explain why units are important when constructing and solving scientific equations.

4. Why might two experimental measurements be insufficient to determine three unknown quantities?

5. Give two assumptions that might limit the accuracy of a scientific mathematical model.

Application: Chemistry and Density

6. A laboratory mixes two liquids.

Liquid A has a density of 1.0 g/mL.

Liquid B has a density of 1.5 g/mL.

The mixture has a total volume of 100 mL and a total mass of 120 g.

Assuming additive volumes, determine the volume of each liquid.

7. A laboratory needs 400 mL of a 25% solution.

It has a 10% solution and a 40% solution.

Assuming the volumes are additive and the concentration percentages are compatible, determine how much of each solution is required.

8. Two substances combine to form 90 g of a compound.

The mass of substance A is twice the mass of substance B.

Find the mass of each substance.

Application: Physics

9. Two vehicles travel in opposite directions from the same starting point.

After 2 hours, they are 140 km apart.

One vehicle travels 10 km/h faster than the other.

Find their speeds.

10. An object moves at constant velocity.

At t = 3 s, its position is 14 m.

At t = 7 s, its position is 30 m.

Use s = vt + s₀ to determine the velocity and initial position.

11. A current of 9 A enters a junction and divides between two parallel resistors.

The resistances are 3 Ω and 6 Ω.

Determine the current through each resistor.

12. Two horizontal forces act on an object.

One force acts to the right with magnitude F₁.

The other acts to the left with magnitude F₂.

The forces have a combined magnitude of 50 N, and the resultant force is 10 N to the right.

Determine F₁ and F₂.

Experimental Modelling

13. A spring experiment follows the model:

F = kx + b

The following measurements are recorded:

Determine k and b, including their units.

14. A scientist models experimental data using:

y = mx + b

Two measured points are:

(2, 9) and (6, 21)

Determine the equation of the line.

15. A laboratory mixture contains three liquids.

Their densities are 1 g/mL, 2 g/mL, and 3 g/mL.

The total volume is 120 mL.

The total mass is 240 g.

The volume of the third liquid is 20 mL greater than the volume of the first.

Determine whether the system has a unique solution.

Analysis and Challenge

16. A scientist calculates that a sample has a mass of −12 g.

Explain why this result should be questioned.

17. An experiment produces the equations:

x + y = 20

2x + 2y = 40

Explain why the unknowns cannot be determined uniquely.

18. A motion model assumes constant velocity. However, an object is observed to accelerate.

Explain why the model may no longer be appropriate.

19. A student solves:

x + y = 10

x + 1.001y = 10.002

and obtains x = 8, y = 2.

Explain why this system may be sensitive to measurement errors.

20. A scientist determines two unknown concentrations using a system of equations. The calculated concentrations satisfy the equations, but one concentration exceeds the maximum physically possible concentration.

Explain what the scientist should conclude and how the investigation might be improved.

Answers to Check Your Understanding

1. Why systems are useful

Systems of equations allow scientists to combine several measurements or scientific relationships to determine multiple unknown quantities.

They are useful in problems involving mixtures, motion, forces, circuits, and experimental data.

2. Mathematical and scientific validity

A mathematically correct solution satisfies the equations.

A scientifically valid solution must also agree with physical laws, appropriate units, realistic conditions, and the assumptions of the model.

3. Importance of units

Units identify the physical quantities represented by variables.

They also help detect incorrect equations and prevent mistakes caused by mixing measurements such as metres and kilometres or litres and millilitres.

4. Insufficient measurements

Three unknown quantities generally require three independent relationships to determine a unique solution in a linear system.

Two measurements may leave infinitely many possible combinations.

Additional independent information may be required.

5. Model assumptions

Examples include:

• Assuming constant velocity.

• Ignoring air resistance.

• Assuming liquid volumes are additive.

• Assuming electrical resistance remains constant.

• Treating experimental measurements as exact.

Any two suitable examples are acceptable.

6. Density mixture

Let:

x = volume of Liquid A, in mL

y = volume of Liquid B, in mL

Total volume:

x + y = 100

Total mass:

1.0x + 1.5y = 120

From the first equation:

x = 100 − y

Substitute:

100 − y + 1.5y = 120

0.5y = 20

y = 40

Then:

x = 60

Answer: 60 mL of Liquid A and 40 mL of Liquid B.

Verification:

60 + 40 = 100 mL

1.0(60) + 1.5(40) = 120 g

7. Solution concentrations

Let:

x = volume of 10% solution

y = volume of 40% solution

Total volume:

x + y = 400

Solute equation:

0.10x + 0.40y = 0.25(400)

0.10x + 0.40y = 100

Substitute x = 400 − y:

0.10(400 − y) + 0.40y = 100

40 − 0.10y + 0.40y = 100

0.30y = 60

y = 200

Then:

x = 200

Answer: 200 mL of each solution.

Verification:

0.10(200) + 0.40(200) = 100

100/400 × 100% = 25%

8. Conservation of mass

Let:

x = mass of substance A

y = mass of substance B

Equations:

x + y = 90

x = 2y

Substitute:

2y + y = 90

3y = 90

y = 30

x = 60

Answer: Substance A = 60 g; substance B = 30 g.

9. Two vehicles

Let:

x = slower speed, in km/h

y = faster speed, in km/h

Distance equation:

2x + 2y = 140

x + y = 70

Speed difference:

y − x = 10

Add:

2y = 80

y = 40

Then:

x = 30

Answer: 30 km/h and 40 km/h.

Verification:

2(30) + 2(40) = 140 km

40 − 30 = 10 km/h

10. Constant-velocity motion

Use:

s = vt + s₀

Measurements:

14 = 3v + s₀

30 = 7v + s₀

Subtract:

16 = 4v

v = 4 m/s

Substitute:

14 = 3(4) + s₀

s₀ = 2 m

Answer: Velocity = 4 m/s; initial position = 2 m.

The model is:

s = 4t + 2

11. Electrical currents

Let:

I₁ = current through 3 Ω resistor

I₂ = current through 6 Ω resistor

Current law:

I₁ + I₂ = 9

Equal voltage across parallel branches:

3I₁ = 6I₂

Therefore:

I₁ = 2I₂

Substitute:

2I₂ + I₂ = 9

3I₂ = 9

I₂ = 3 A

I₁ = 6 A

Answer: 6 A through the 3 Ω resistor and 3 A through the 6 Ω resistor.

Verification:

Total current = 9 A

Voltage across both branches = 18 V

12. Forces in one dimension

Let:

F₁ = rightward force magnitude

F₂ = leftward force magnitude

Combined magnitude:

F₁ + F₂ = 50

Resultant force:

F₁ − F₂ = 10

Add:

2F₁ = 60

F₁ = 30 N

Then:

F₂ = 20 N

Answer: 30 N to the right and 20 N to the left.

Resultant:

30 − 20 = 10 N to the right.

13. Spring constant and offset

Model:

F = kx + b

Measurements:

4 = 0.03k + b

8 = 0.07k + b

Subtract:

4 = 0.04k

k = 100 N/m

Substitute:

4 = 0.03(100) + b

4 = 3 + b

b = 1 N

Answer: k = 100 N/m; b = 1 N.

Model:

F = 100x + 1

with F in N and x in m.

14. Linear experimental model

Given:

9 = 2m + b

21 = 6m + b

Subtract:

12 = 4m

m = 3

Substitute:

9 = 2(3) + b

b = 3

Answer: y = 3x + 3.

The units of the gradient and intercept depend on the quantities represented by x and y.

15. Three-liquid mixture

Let:

x = volume of first liquid

y = volume of second liquid

z = volume of third liquid

Equations:

x + y + z = 120

x + 2y + 3z = 240

z − x = 20

Subtract the volume equation from the mass equation:

y + 2z = 120

From the third equation:

x = z − 20

Substitute into the volume equation:

z − 20 + y + z = 120

y + 2z = 140

This contradicts the earlier equation:

y + 2z = 120

Answer: The system has no solution.

The stated measurements and volume relationship are inconsistent under the model's assumptions.

The scientist should check the data, units, and assumptions before proceeding.

16. Negative mass

Mass cannot be negative in this context.

The scientist should check the equations, measurements, units, and calculations.

The negative value may indicate an incorrect model or inconsistent experimental data.

17. Dependent equations

The second equation is twice the first.

Therefore, it does not provide new information.

The system has infinitely many solutions:

x = t

y = 20 − t

where t is any real number.

An additional independent equation is needed to determine unique values.

18. Constant-velocity model

The equation:

s = vt + s₀

assumes constant velocity.

If the object accelerates, velocity changes with time.

For constant acceleration, a more appropriate model may be:

s = s₀ + ut + ½at²

where u is initial velocity and a is constant acceleration.

For changing acceleration, a more complex model may be required.

19. Sensitivity to measurement errors

The equations are:

x + y = 10

x + 1.001y = 10.002

Subtract:

0.001y = 0.002

y = 2

x = 8

The coefficient difference is very small.

Therefore, even a small change in the measured constant can produce a large change in the calculated variables.

The system is ill-conditioned.

This means the solution may be unreliable when measurements contain uncertainty.

20. Physically impossible concentration

The calculated solution may satisfy the mathematical equations but violate physical constraints.

The scientist should not automatically accept it as scientifically valid.

Possible actions include checking the original measurements, confirming units and concentration definitions, examining the model assumptions, and collecting additional measurements.

The scientist should also investigate whether the system is sensitive to measurement errors.

A revised model or improved experimental procedure may be necessary.

Final Summary

Systems of equations are valuable tools for analysing scientific situations involving several unknown quantities. They allow measurements and scientific laws to be combined into mathematical models that can be solved using algebraic methods or technology.

Applications include chemical mixtures, concentration calculations, motion, force equilibrium, electrical circuits, biological populations, and experimental data analysis.

However, scientific problem-solving requires more than obtaining numerical answers. Results must be checked using correct units, physical laws, realistic constraints, and the assumptions of the model.

The strongest scientific solutions combine accurate mathematics with careful measurement, sound scientific reasoning, and a clear understanding of the model's limitations.

3. Geometry Applications

Introduction

Geometry is the study of shapes, sizes, positions, distances, angles, and spatial relationships. Many geometric problems involve two or more unknown quantities connected by mathematical rules.

For example, we might need to determine:

• The length and width of a rectangle.

• The angles of a triangle.

• The coordinates where two lines intersect.

• The dimensions of a shape from its perimeter and area.

• The point where a line intersects a circle.

• The coordinates of the vertices of a geometric figure.

• The dimensions of a three-dimensional object.

Systems of equations provide a systematic way to solve these problems by combining different geometric relationships.

In coordinate geometry, equations can represent lines, circles, and other curves. Solving a system allows us to identify points shared by two or more geometric objects.

In measurement problems, equations can describe relationships between lengths, widths, heights, perimeters, areas, and volumes.

By constructing appropriate systems, solving them, and interpreting the results geometrically, we can solve problems that would otherwise be difficult to approach.

Why Are Systems Useful in Geometry?

A single geometric condition often does not provide enough information to determine every unknown.

Suppose a rectangle has a perimeter of 40 cm.

Its length and width satisfy:

2l + 2w = 40

or:

l + w = 20

Many rectangles satisfy this condition.

Examples include:

• Length 12 cm and width 8 cm.

• Length 15 cm and width 5 cm.

• Length 11 cm and width 9 cm.

The perimeter alone does not determine a unique rectangle.

However, suppose we also know that the length is 4 cm greater than the width.

Then:

l − w = 4

We now have a system:

l + w = 20

l − w = 4

Solving this system determines the dimensions uniquely.

This illustrates an important idea:

Each independent geometric condition provides information that can help determine unknown quantities.

Constructing Systems from Geometric Relationships

A reliable approach to geometric modelling involves six steps.

1. Identify the unknown lengths, angles, or coordinates.

2. Define variables and include units where appropriate.

3. Identify relevant geometric properties or formulas.

4. Construct equations from the given conditions.

5. Solve the equations and check all possible solutions.

6. Verify that the results satisfy both the equations and the geometric restrictions.

Useful geometric relationships

Some geometric relationships produce linear equations, while others produce nonlinear equations involving squares or products of variables.

Systems of equations can be used in both situations, although the solving methods may differ.

Application: Finding the Dimensions of a Rectangle

A rectangular garden has a perimeter of 52 m.

Its length is 6 m greater than its width.

Find the dimensions.

Step 1: Define variables

l = length of the garden, in metres

w = width of the garden, in metres

Step 2: Construct the perimeter equation

P = 2l + 2w

52 = 2l + 2w

Divide by 2:

l + w = 26

Step 3: Construct the second equation

The length is 6 m greater than the width.

l − w = 6

Step 4: Solve by elimination

l + w = 26

l − w = 6

Add:

2l = 32

l = 16

Substitute:

16 + w = 26

w = 10

Step 5: Verify

Perimeter:

2(16) + 2(10) = 52 m

Difference:

16 − 10 = 6 m

Both conditions are satisfied.

Answer: The garden is 16 m long and 10 m wide.

Geometric interpretation

The solution determines a unique rectangle satisfying both the perimeter and dimension conditions.

Its area is:

A = lw

A = 16(10)

A = 160 m²

The system provides dimensions that can be used for further calculations involving fencing, landscaping, or construction materials.

Application: Rectangle Dimensions from Area and Perimeter

Not all geometric systems are linear.

Suppose a rectangle has:

• Perimeter = 34 cm.

• Area = 60 cm².

Find its dimensions.

Step 1: Define variables

l = length, in cm

w = width, in cm

Step 2: Construct equations

Perimeter:

2l + 2w = 34

Simplify:

l + w = 17

Area:

lw = 60

Step 3: Substitute

From the perimeter equation:

l = 17 − w

Substitute into the area equation:

(17 − w)w = 60

17w − w² = 60

Rearrange:

w² − 17w + 60 = 0

Step 4: Factor

(w − 5)(w − 12) = 0

Therefore:

w = 5 or w = 12

If w = 5:

l = 12

If w = 12:

l = 5

These describe the same rectangle with its dimensions interchanged.

Answer: The rectangle measures 12 cm by 5 cm.

Verify

Perimeter:

2(12) + 2(5) = 34 cm

Area:

12(5) = 60 cm²

Important observation

The system contains a linear equation and a nonlinear equation.

Substitution transforms the system into a quadratic equation.

This demonstrates how systems of equations connect algebra and geometry.

Application: Finding Unknown Angles

Geometric angle relationships can be represented using systems of equations.

Worked Example: Complementary Angles

Two angles are complementary.

The larger angle is 18° greater than the smaller angle.

Find both angles.

Step 1: Define variables

x = larger angle, in degrees

y = smaller angle, in degrees

Step 2: Construct equations

Complementary angles add to 90°.

x + y = 90

The larger angle exceeds the smaller by 18°.

x − y = 18

Step 3: Solve

Add the equations:

2x = 108

x = 54

Then:

y = 90 − 54

y = 36

Step 4: Verify

54° + 36° = 90°

54° − 36° = 18°

Answer: The angles are 54° and 36°.

Application: Angles in a Triangle

The three interior angles of a triangle add to 180°.

Suppose:

• The second angle is 20° greater than the first.

• The third angle is twice the first.

Find all three angles.

Step 1: Define variables

x = first angle

y = second angle

z = third angle

Step 2: Construct equations

Angle sum:

x + y + z = 180

Second angle relationship:

y = x + 20

Third angle relationship:

z = 2x

Step 3: Substitute

x + (x + 20) + 2x = 180

4x + 20 = 180

4x = 160

x = 40

Then:

y = 40 + 20 = 60

z = 2(40) = 80

Step 4: Verify

40° + 60° + 80° = 180°

60° − 40° = 20°

80° = 2(40°)

Answer: The triangle's angles are 40°, 60°, and 80°.

Geometric interpretation

All three angles are less than 90°, so the triangle is acute.

This example shows how a three-variable system can be solved efficiently using relationships between the variables.

Coordinate Geometry: Intersections of Straight Lines

In coordinate geometry, a straight line can be represented by an equation such as:

y = mx + b

where:

• m = gradient or slope.

• b = y-intercept.

• x and y = coordinates of points on the line.

When two lines intersect, the intersection point satisfies both equations simultaneously.

Therefore, finding the intersection of two lines is equivalent to solving a system of equations.

Worked Example: Two Intersecting Lines

Find the intersection of:

y = 2x + 1

y = −x + 7

Step 1: Set the expressions equal

At the intersection, both equations have the same x-value and y-value.

2x + 1 = −x + 7

Step 2: Solve for x

3x = 6

x = 2

Step 3: Solve for y

y = 2(2) + 1

y = 5

Intersection: (2, 5).

Step 4: Verify

First equation:

5 = 2(2) + 1

Second equation:

5 = −2 + 7

Both are correct.

Geometric interpretation

The ordered pair (2, 5) represents a point that lies on both lines.

The x-coordinate is 2, and the y-coordinate is 5.

Because the lines have different gradients, they intersect at exactly one point.

Parallel and Coincident Lines

Not every pair of lines has one intersection.

Consider three possibilities.

Example: Parallel Lines

y = 3x + 2

y = 3x − 4

Setting the expressions equal:

3x + 2 = 3x − 4

2 = −4

This is impossible.

Conclusion: No solution.

Example: Coincident Lines

2x + 4y = 12

x + 2y = 6

Multiplying the second equation by 2 produces the first equation.

The equations represent the same line.

Conclusion: Infinitely many solutions.

Every point on that line satisfies both equations.

Application: Finding the Vertices of a Triangle

A triangle is formed by three lines:

Line A: y = x + 1

Line B: y = −x + 7

Line C: y = 2

Find the three vertices of the triangle.

Step 1: Find the intersection of A and B

x + 1 = −x + 7

2x = 6

x = 3

y = 4

First vertex:

P = (3, 4)

Step 2: Find the intersection of A and C

2 = x + 1

x = 1

Second vertex:

Q = (1, 2)

Step 3: Find the intersection of B and C

2 = −x + 7

x = 5

Third vertex:

R = (5, 2)

Step 4: Interpret geometrically

The vertices are:

P = (3, 4)

Q = (1, 2)

R = (5, 2)

Step 5: Calculate the area

The base QR lies horizontally along y = 2.

Its length is:

5 − 1 = 4 units

The vertical height from P to the base is:

4 − 2 = 2 units

Area of a triangle:

A = ½bh

A = ½(4)(2)

A = 4 square units

Verification

Check each vertex in its two corresponding line equations.

For P = (3, 4):

4 = 3 + 1

4 = −3 + 7

For Q = (1, 2):

2 = 1 + 1

2 = 2

For R = (5, 2):

2 = −5 + 7

2 = 2

All three vertices are correct.

Application: Finding the Intersection of a Line and a Circle

Systems can also contain nonlinear equations.

The equation of a circle with centre (h, k) and radius r is:

(x − h)² + (y − k)² = r²

Suppose we want to find where the line:

y = x + 1

intersects the circle:

x² + y² = 25

Step 1: Substitute the line equation

Replace y with x + 1:

x² + (x + 1)² = 25

Step 2: Expand

x² + x² + 2x + 1 = 25

2x² + 2x − 24 = 0

Divide by 2:

x² + x − 12 = 0

Step 3: Factor

(x + 4)(x − 3) = 0

Therefore:

x = −4 or x = 3

Step 4: Find the corresponding y-values

If x = −4:

y = −4 + 1 = −3

If x = 3:

y = 3 + 1 = 4

Intersection points: (−4, −3) and (3, 4).

Step 5: Verify

For (−4, −3):

Circle:

(−4)² + (−3)² = 16 + 9 = 25

Line:

−3 = −4 + 1

For (3, 4):

Circle:

3² + 4² = 9 + 16 = 25

Line:

4 = 3 + 1

Both points satisfy both equations.

How many intersections are possible?

A straight line and a circle can have:

• Two intersection points.

• One intersection point, when the line is tangent.

• No real intersection points.

The number of real solutions of the resulting quadratic equation determines the number of intersections.

Using the discriminant

For:

ax² + bx + c = 0

the discriminant is:

D = b² − 4ac

If:

• D is positive: two distinct real intersections.

• D = 0: one tangent intersection.

• D is negative: no real intersections.

This provides a connection between quadratic equations and coordinate geometry.

Application: Finding an Unknown Point Using the Midpoint

The midpoint of a line segment lies halfway between its endpoints.

For endpoints A(x₁, y₁) and B(x₂, y₂), the midpoint is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Suppose a line segment has:

A = (2, 4)

M = (5, 7)

Find the coordinates of B.

Step 1: Define the unknown endpoint

B = (x, y)

Step 2: Construct equations

For the x-coordinate:

(2 + x)/2 = 5

For the y-coordinate:

(4 + y)/2 = 7

Step 3: Solve

2 + x = 10

x = 8

4 + y = 14

y = 10

Answer: B = (8, 10).

Verify

Midpoint of (2, 4) and (8, 10):

M = ((2 + 8)/2, (4 + 10)/2)

M = (5, 7)

The calculated endpoint is correct.

Geometric interpretation

The midpoint is exactly halfway between the endpoints in both coordinate directions.

The two coordinate equations can be solved independently because each involves a different unknown.

Application: Perpendicular Lines and Intersection Points

Two nonvertical lines are perpendicular when their gradients multiply to −1.

m₁m₂ = −1

Suppose a line passes through (2, 3) and has gradient 2.

Find the equation of the perpendicular line passing through (4, 1), and determine where the two lines intersect.

Step 1: Find the first line

Using point-slope form:

y − 3 = 2(x − 2)

y = 2x − 1

Step 2: Find the perpendicular gradient

The negative reciprocal of 2 is:

m = −1/2

Step 3: Find the second line

Through (4, 1):

y − 1 = −½(x − 4)

y = −½x + 3

Step 4: Solve the system

2x − 1 = −½x + 3

2.5x = 4

x = 8/5

Step 5: Find y

y = 2(8/5) − 1

y = 16/5 − 5/5

y = 11/5

Intersection: (8/5, 11/5).

Verify

First line:

2(8/5) − 1 = 11/5

Second line:

−½(8/5) + 3 = −4/5 + 15/5 = 11/5

Gradient product:

2(−1/2) = −1

Both the intersection and perpendicularity conditions are satisfied.

Geometric interpretation

The two lines meet at a right angle.

This method is useful when constructing perpendicular paths, locating perpendicular projections, or analysing geometric figures.

Application: Dimensions of a Rectangular Prism

Three-dimensional geometry can involve three unknown dimensions.

A rectangular storage box has:

• Length + width + height = 24 cm.

• Length is twice the width.

• Height is 3 cm greater than the width.

Find the dimensions and volume.

Step 1: Define variables

l = length, in cm

w = width, in cm

h = height, in cm

Step 2: Construct equations

l + w + h = 24

l = 2w

h = w + 3

Step 3: Substitute

2w + w + (w + 3) = 24

4w + 3 = 24

4w = 21

w = 21/4 = 5.25 cm

Step 4: Find the remaining dimensions

l = 2(5.25) = 10.5 cm

h = 5.25 + 3 = 8.25 cm

Step 5: Verify

10.5 + 5.25 + 8.25 = 24 cm

10.5 = 2(5.25)

8.25 = 5.25 + 3

All conditions are satisfied.

Step 6: Calculate volume

V = lwh

V = 10.5 × 5.25 × 8.25

V = 454.78125 cm³

Interpretation

The box measures:

• Length: 10.5 cm.

• Width: 5.25 cm.

• Height: 8.25 cm.

Its volume is approximately 454.78 cm³.

This type of modelling is useful in packaging, storage design, and construction.

Applications in Design, Architecture, and Engineering

Systems of equations are used extensively when geometric conditions must be satisfied simultaneously.

Example: Locating Two Paths

Two straight paths are modelled by:

Path A: y = 0.5x + 2

Path B: y = −x + 8

Their intersection is found by solving:

0.5x + 2 = −x + 8

1.5x = 6

x = 4

Then:

y = 0.5(4) + 2

y = 4

The paths intersect at (4, 4).

If the coordinates are measured in metres from a defined origin, the intersection is 4 m east and 4 m north of that origin, assuming positive x points east and positive y points north.

In a real design, the coordinate system and measurement units must be clearly defined.

Using Technology to Verify Geometric Solutions

Graphing calculators and mathematical software can help visualise geometric systems.

For example, enter:

y = 2x + 1

y = −x + 7

into Desmos Graphing Calculator.

The graph shows the lines intersecting at (2, 5).

For a line-circle system, enter:

x² + y² = 25

y = x + 1

The graph displays two intersection points.

Benefits of technology

Technology can:

• Display geometric relationships visually.

• Estimate intersection coordinates.

• Check algebraic solutions.

• Investigate the effect of changing parameters.

• Help identify the number of intersections.

Limitations

Graphing technology may display approximate values.

A poorly chosen graphing window may hide intersections.

Some intersections may be difficult to identify accurately from the graph alone.

Therefore, algebraic verification remains important.

Evaluating Geometric Solutions

A solution may satisfy an equation but still be inappropriate for the geometric situation.

Positive dimensions

Lengths, widths, heights, radii, and areas cannot be negative in ordinary geometric measurement.

For example, a quadratic equation might produce:

w = 5 or w = −3

If w represents a physical width, the negative solution must be rejected.

Angle restrictions

Interior angles of a nondegenerate triangle must be greater than 0° and less than 180°.

Three proposed angles must also add to 180°.

Coordinate order

The point (3, 5) is different from (5, 3).

Coordinate order must be preserved.

Extraneous solutions

Squaring equations or performing other non-equivalent transformations may introduce solutions that do not satisfy the original system.

Every candidate solution should be checked.

Geometric assumptions

A drawing may not be to scale.

Do not assume two lines are perpendicular or two lengths are equal unless this is given or established mathematically.

Units

Lengths are measured in units such as cm or m.

Areas are measured in square units such as cm² or m².

Volumes are measured in cubic units such as cm³ or m³.

Using incorrect units can make an otherwise correct numerical result incomplete.

Common Mistakes

Using the wrong perimeter formula

For a rectangle:

P = 2l + 2w

not l + w.

Confusing area and perimeter

Area measures the region enclosed by a shape, while perimeter measures the distance around it.

Forgetting angle-sum relationships

Triangle angles add to 180°, while quadrilateral angles add to 360°.

Reversing coordinate values

The point (x, y) must be written in the correct order.

Assuming every system is linear

Area and circle equations often produce nonlinear systems.

Ignoring multiple solutions

A line may intersect a circle at two points.

Both points must be included when valid.

Accepting negative dimensions

Negative lengths or widths are not valid physical dimensions.

Assuming a diagram is drawn to scale

Only use relationships given or justified by geometry.

Misinterpreting parallel lines

Parallel lines with different intercepts have no intersection.

Forgetting to verify intersections

A coordinate must satisfy both equations to be an intersection point.

Using incorrect units

Lengths, areas, and volumes require different types of units.

Relying only on graphing technology

Graphical answers may be approximate or incomplete.

Key Terms

• System of equations: Two or more equations solved simultaneously.

• Coordinate geometry: The study of geometric figures using coordinates and equations.

• Cartesian plane: A coordinate plane with perpendicular x- and y-axes.

• Ordered pair: A pair of coordinates written as (x, y).

• Intersection point: A point shared by two or more geometric objects.

• Gradient: The slope of a straight line.

• Y-intercept: The point where a line crosses the y-axis.

• Parallel lines: Coplanar lines that never intersect.

• Perpendicular lines: Lines that meet at a right angle.

• Coincident lines: Lines that occupy exactly the same position.

• Perimeter: The total distance around a shape.

• Area: The amount of two-dimensional space enclosed by a shape.

• Volume: The amount of three-dimensional space occupied by an object.

• Complementary angles: Two angles whose sum is 90°.

• Supplementary angles: Two angles whose sum is 180°.

• Midpoint: The point halfway between two endpoints.

• Circle: The set of points at a fixed distance from a centre.

• Radius: The distance from the centre of a circle to its boundary.

• Tangent: A line that touches a circle at exactly one point.

• Discriminant: The expression b² − 4ac, which helps determine the number of real roots of a quadratic equation.

• Nonlinear system: A system containing at least one nonlinear equation.

• Extraneous solution: A solution introduced during algebraic manipulation that does not satisfy the original problem.

• Geometric constraint: A condition restricting possible dimensions, angles, or positions.

• Verification: Checking that a solution satisfies the original equations and geometric conditions.

Key Takeaways

• Systems of equations can model lengths, widths, angles, coordinates, and dimensions.

• Geometric properties provide relationships that can be expressed as equations.

• Rectangle problems may involve perimeter, area, and dimension relationships.

• Angle problems can use triangle angle sums and relationships between angles.

• The intersection of two lines is a solution common to both line equations.

• Parallel lines have no intersection, while coincident lines have infinitely many shared points.

• Triangle vertices can be found by solving pairs of line equations.

• A line and a circle may have zero, one, or two real intersection points.

• Midpoint relationships can be used to find unknown coordinates.

• Three-variable systems can determine the dimensions of three-dimensional objects.

• Geometric solutions must satisfy relevant restrictions and use correct units.

• Verification is essential for confirming that calculated values represent valid geometric results.

Check Your Understanding

Recall and Understanding

1. Explain how a system of equations can be used to determine the dimensions of a rectangle.

2. What does the solution to a system of two straight-line equations represent geometrically?

3. Explain the difference between intersecting, parallel, and coincident lines.

4. Why might a line-circle system have two solutions?

5. Why must negative solutions sometimes be rejected in geometric problems?

Rectangle and Angle Problems

6. A rectangle has a perimeter of 48 cm.

Its length is 8 cm greater than its width.

Find its dimensions and area.

7. A rectangle has a perimeter of 30 cm and an area of 54 cm².

Find its dimensions.

8. Two complementary angles differ by 24°.

Find both angles.

9. The second angle of a triangle is 15° greater than the first.

The third angle is twice the first.

Find all three angles.

10. Three angles of a quadrilateral are x, x + 20°, and 2x. The fourth angle is 100°.

Find all four angles.

Coordinate Geometry

11. Find the intersection of:

y = 3x − 2

y = −x + 10

12. Determine the number of solutions:

y = 4x + 3

y = 4x − 7

13. Determine the number of solutions:

2x + 6y = 18

x + 3y = 9

14. Find the vertices of the triangle formed by:

y = x

y = −x + 6

y = 1

15. Find the intersection points of:

x² + y² = 25

y = 3

Applications and Challenge

16. A line segment has endpoint A = (3, 5) and midpoint M = (7, 9).

Find the other endpoint.

17. A line has gradient 3 and passes through (1, 2).

Find the equation of the perpendicular line passing through (4, 5), then determine the intersection of the two lines.

18. A rectangular prism has length, width, and height whose sum is 30 cm.

The length is twice the width.

The height is 6 cm greater than the width.

Find its dimensions and volume.

19. Find the intersection points of:

y = x + 2

x² + y² = 20

20. A rectangular garden has a perimeter of 60 m and an area of 200 m².

Find its dimensions and explain how you verify the result.

Answers to Check Your Understanding

1. Rectangle dimensions

Define variables for length and width.

Use geometric information, such as perimeter and a relationship between the dimensions, to construct two equations.

Solve the system and verify that the dimensions satisfy the original conditions.

2. Intersection of two lines

The solution represents the coordinates of a point lying on both lines.

For example, (3, 5) means both lines pass through x = 3, y = 5.

3. Types of lines

Intersecting lines cross at one point and have one solution.

Distinct parallel lines never meet and have no solution.

Coincident lines represent the same line and have infinitely many solutions.

4. Line-circle intersections

A line can pass through a circle at two different points.

Both intersection points satisfy the line equation and circle equation.

The resulting quadratic equation may therefore have two distinct real solutions.

5. Negative geometric values

Physical dimensions such as lengths, widths, and radii cannot be negative.

Therefore, a negative algebraic solution must be rejected when it represents an invalid dimension.

Negative coordinates, however, can be perfectly valid.

6. Rectangle perimeter

Let:

l = length

w = width

Perimeter:

2l + 2w = 48

l + w = 24

Difference:

l − w = 8

Add:

2l = 32

l = 16

Then:

w = 8

Area:

A = 16(8) = 128 cm²

Answer: Length = 16 cm, width = 8 cm, area = 128 cm².

7. Rectangle area and perimeter

Perimeter:

2l + 2w = 30

l + w = 15

Area:

lw = 54

Substitute l = 15 − w:

(15 − w)w = 54

15w − w² = 54

w² − 15w + 54 = 0

(w − 6)(w − 9) = 0

Therefore:

w = 6 or w = 9

Answer: The rectangle measures 9 cm by 6 cm.

Verification:

2(9) + 2(6) = 30 cm

9(6) = 54 cm²

8. Complementary angles

Let:

x = larger angle

y = smaller angle

Equations:

x + y = 90

x − y = 24

Add:

2x = 114

x = 57

Then:

y = 33

Answer: 57° and 33°.

9. Triangle angles

Let the first angle be x.

Second angle:

x + 15

Third angle:

2x

Angle sum:

x + (x + 15) + 2x = 180

4x + 15 = 180

4x = 165

x = 41.25

Second angle:

56.25°

Third angle:

82.5°

Answer: 41.25°, 56.25°, and 82.5°.

Verification:

41.25 + 56.25 + 82.5 = 180°

10. Quadrilateral angles

Angle sum:

x + (x + 20) + 2x + 100 = 360

4x + 120 = 360

4x = 240

x = 60

Therefore:

First angle = 60°

Second angle = 80°

Third angle = 120°

Fourth angle = 100°

Answer: 60°, 80°, 120°, and 100°.

Verification:

60 + 80 + 120 + 100 = 360°

11. Intersection of two lines

Equations:

y = 3x − 2

y = −x + 10

Set equal:

3x − 2 = −x + 10

4x = 12

x = 3

Then:

y = 3(3) − 2 = 7

Answer: (3, 7).

Verification:

7 = −3 + 10

12. Parallel lines

Equations:

y = 4x + 3

y = 4x − 7

Both lines have gradient 4 but different y-intercepts.

Answer: No solution.

13. Coincident lines

Equations:

2x + 6y = 18

x + 3y = 9

Multiplying the second equation by 2 gives the first.

Answer: Infinitely many solutions.

14. Triangle vertices

Lines:

y = x

y = −x + 6

y = 1

First intersection:

x = −x + 6

2x = 6

x = 3

y = 3

First vertex: (3, 3)

Second intersection:

1 = x

Second vertex: (1, 1)

Third intersection:

1 = −x + 6

x = 5

Third vertex: (5, 1)

Answer: (3, 3), (1, 1), and (5, 1).

15. Line-circle intersections

Circle:

x² + y² = 25

Line:

y = 3

Substitute:

x² + 3² = 25

x² + 9 = 25

x² = 16

x = ±4

Answer: (−4, 3) and (4, 3).

Verification:

16 + 9 = 25 for both points.

16. Unknown endpoint

Given:

A = (3, 5)

M = (7, 9)

Let the other endpoint be B = (x, y).

Midpoint equations:

(3 + x)/2 = 7

(5 + y)/2 = 9

Solve:

3 + x = 14

x = 11

5 + y = 18

y = 13

Answer: B = (11, 13).

17. Perpendicular lines

The first line has gradient 3 and passes through (1, 2).

Using point-slope form:

y − 2 = 3(x − 1)

y = 3x − 1

The perpendicular gradient is:

−1/3

Through (4, 5):

y − 5 = −⅓(x − 4)

y = −x/3 + 19/3

Find the intersection:

3x − 1 = −x/3 + 19/3

Multiply by 3:

9x − 3 = −x + 19

10x = 22

x = 11/5

Substitute:

y = 3(11/5) − 1

y = 28/5

Answer: The perpendicular line is y = −x/3 + 19/3, and the intersection is (11/5, 28/5).

Verification:

3(−1/3) = −1

The gradients confirm that the lines are perpendicular.

18. Rectangular prism

Let:

l = length

w = width

h = height

Equations:

l + w + h = 30

l = 2w

h = w + 6

Substitute:

2w + w + (w + 6) = 30

4w + 6 = 30

4w = 24

w = 6

Then:

l = 12

h = 12

Volume:

V = lwh

V = 12(6)(12)

V = 864 cm³

Answer: Length = 12 cm, width = 6 cm, height = 12 cm, volume = 864 cm³.

19. Line-circle intersections

Given:

y = x + 2

x² + y² = 20

Substitute:

x² + (x + 2)² = 20

Expand:

x² + x² + 4x + 4 = 20

2x² + 4x − 16 = 0

Divide by 2:

x² + 2x − 8 = 0

Factor:

(x + 4)(x − 2) = 0

Therefore:

x = −4 or x = 2

If x = −4:

y = −2

If x = 2:

y = 4

Answer: (−4, −2) and (2, 4).

Verification:

For (−4, −2):

16 + 4 = 20

For (2, 4):

4 + 16 = 20

Both points also satisfy y = x + 2.

20. Rectangular garden

Given:

Perimeter = 60 m

Area = 200 m²

Let:

l = length

w = width

Perimeter:

2l + 2w = 60

l + w = 30

Area:

lw = 200

Substitute:

l = 30 − w

(30 − w)w = 200

30w − w² = 200

w² − 30w + 200 = 0

Factor:

(w − 10)(w − 20) = 0

Therefore:

w = 10 or w = 20

The dimensions are interchangeable.

Answer: The garden measures 20 m by 10 m.

Verification:

Perimeter:

2(20) + 2(10) = 60 m

Area:

20(10) = 200 m²

Both original conditions are satisfied.

Final Summary

Systems of equations provide a powerful connection between algebra and geometry. They allow us to determine unknown dimensions, calculate angles, find intersections, locate vertices, and analyse geometric figures in two and three dimensions.

Some geometric systems are linear, while others involve quadratic equations or additional geometric relationships.

The most important part of solving these problems is constructing equations that accurately represent the geometric conditions.

Once a solution has been obtained, it should be verified using the original equations, relevant geometric properties, and appropriate units.

Successful geometric problem-solving combines algebraic accuracy, visual understanding, and careful interpretation of mathematical results.

4. Mathematical Modelling

Introduction

Mathematics is more than a collection of equations and calculations. It is a powerful tool for understanding the world, making predictions, solving practical problems, and supporting decisions.

Scientists use mathematics to describe motion and chemical reactions. Businesses use it to analyse costs and profits. Engineers use it to design structures. Environmental researchers use it to investigate pollution, energy consumption, and resource use.

In many of these situations, several unknown quantities are connected by different relationships. Systems of equations allow us to represent these relationships mathematically and investigate how they interact.

The process of converting a real-world situation into mathematical relationships is called mathematical modelling.

A mathematical model is not necessarily an exact description of reality. Instead, it is a simplified representation that focuses on the most important features of a situation.

A good mathematical model should be:

• Based on relevant information.

• Mathematically consistent.

• Appropriate for the situation.

• Clear about its assumptions.

• Useful for answering a particular question.

• Tested against real-world evidence when possible.

In this topic, we will explore how to construct systems-based models, solve them, interpret their results, and evaluate how well they represent reality.

What Is Mathematical Modelling?

Mathematical modelling is the process of representing a real-world situation using mathematical ideas, relationships, and equations.

A model may involve:

• Variables representing unknown quantities.

• Equations describing relationships.

• Constants representing known values.

• Constraints restricting possible solutions.

• Assumptions that simplify the situation.

For example, suppose a company manufactures two products.

Let:

x = number of Product A units.

y = number of Product B units.

The company produces 200 units in total.

This gives:

x + y = 200

Product A requires 2 hours of labour, while Product B requires 3 hours.

The company uses 480 labour-hours.

This gives:

2x + 3y = 480

Together, these equations form a mathematical model of the production situation.

The system is:

x + y = 200

2x + 3y = 480

Solving the system determines the production quantities.

Solving the introductory model

Multiply the first equation by 2:

2x + 2y = 400

Subtract this from the second equation:

y = 80

Substitute:

x + 80 = 200

x = 120

Answer: The company manufactures 120 units of Product A and 80 units of Product B.

Verification:

120 + 80 = 200 units

2(120) + 3(80) = 480 labour-hours

The model is mathematically consistent.

However, its usefulness depends on whether the labour requirements and production totals accurately represent the real situation.

The Mathematical Modelling Cycle

Mathematical modelling is not simply a process of writing equations and calculating answers.

It is a cycle involving interpretation, testing, and improvement.

5. Review and Integrated Problem Solving

Introduction

Systems of equations are among the most useful tools in algebra because they allow us to investigate situations involving several unknown quantities and multiple relationships.

Throughout this unit, we have explored how to solve systems using:

• Graphing

• Substitution

• Elimination

• Matrices and technology

• Algebraic manipulation of nonlinear equations

• Mathematical modelling

We have also applied systems to geometry, science, business, economics, and everyday problems.

The challenge in advanced problem-solving is not simply knowing how to use these methods. It is knowing which method to choose, why it is appropriate, and how to apply it accurately.

A skilled problem-solver can examine a system, identify its structure, select an efficient strategy, complete the necessary calculations, and explain what the solution means.

In unfamiliar problems, the equations may not be given directly. We may need to define variables, interpret information, construct relationships, and combine several mathematical ideas before solving.

This lesson brings these skills together through a comprehensive review of methods, multi-step worked examples, error analysis, and integrated applications.

Understanding What It Means to Solve a System

A system of equations consists of two or more equations involving shared unknown quantities.

A solution must satisfy every equation in the system simultaneously.

For example:

2x + y = 11

x − y = 1

Adding the equations:

3x = 12

x = 4

Substituting into the second equation:

4 − y = 1

y = 3

The solution is:

(x, y) = (4, 3)

Verification

First equation:

2(4) + 3 = 11

Second equation:

4 − 3 = 1

Both are satisfied.

A complete solution should show not only the final values but also the mathematical reasoning used to obtain them.

The Main Methods for Solving Systems

Graphing

Graphing represents each equation visually.

For a system of two straight lines, the intersection point gives the solution.

Best used when:

• A visual interpretation is important.

• The equations are easy to graph.

• An approximate solution is acceptable.

• The number of intersections needs to be investigated.

Limitation: Exact coordinates may be difficult to determine from a graph.

Substitution

Substitution involves isolating one variable and replacing it in another equation.

Best used when:

• One variable is already isolated.

• A coefficient is 1 or −1.

• One equation can be rearranged easily.

• A nonlinear equation is combined with a linear equation.

Elimination

Elimination involves adding or subtracting equations to remove a variable.

Best used when:

• Coefficients are equal or opposite.

• Coefficients can be matched using small multipliers.

• Both equations are in standard form.

• Fractions can be avoided through suitable multiplication.

Matrix methods

Matrix methods represent systems using coefficients and constants.

Best used when:

• There are three or more unknowns.

• Many equations must be solved.

• The same coefficient structure is used repeatedly.

• Technology is available.

Technology

Graphing calculators, spreadsheets, and mathematical software can solve complex systems efficiently.

However, technology should support mathematical reasoning rather than replace it.

Choosing the Most Efficient Method

There is no single best method for every system.

The most efficient choice depends on the structure of the equations.

 

A practical decision guide

Worked Example: Comparing Substitution and Elimination

Solve:

3x + 2y = 18

x + y = 7

Method 1: Substitution

From the second equation:

y = 7 − x

Substitute into the first:

3x + 2(7 − x) = 18

3x + 14 − 2x = 18

x = 4

Then:

y = 7 − 4 = 3

Solution: (4, 3)

Method 2: Elimination

Multiply the second equation by 2:

2x + 2y = 14

Subtract from the first:

x = 4

Then:

4 + y = 7

y = 3

Solution: (4, 3)

Which method is more efficient?

Both methods require only a few steps.

Elimination avoids expanding brackets, while substitution uses the easily isolated variable.

Either is a reasonable choice.

Verification

3(4) + 2(3) = 18

4 + 3 = 7

Both equations are satisfied.

Important principle: Efficiency is not always about using the fewest written lines. A method that is easier to understand and less likely to produce errors may be the better choice.

Worked Example: Elimination with Different Coefficients

Solve:

3x + 4y = 25

5x − 2y = 7

Step 1: Identify the variable to eliminate

The y-coefficients are 4 and −2.

Multiplying the second equation by 2 creates opposite coefficients.

10x − 4y = 14

Step 2: Add the equations

3x + 4y = 25

10x − 4y = 14

Add:

13x = 39

x = 3

Step 3: Find y

Substitute into:

5x − 2y = 7

5(3) − 2y = 7

15 − 2y = 7

−2y = −8

y = 4

Step 4: Verify

3(3) + 4(4) = 9 + 16 = 25

5(3) − 2(4) = 15 − 8 = 7

Answer: (3, 4)

Why elimination is efficient

Only one equation needs to be multiplied.

The y-terms then cancel immediately, producing a simple equation in x.

Worked Example: A System Containing Fractions

Solve:

x/2 + y/3 = 5

x/4 − y/6 = 1

Step 1: Eliminate fractions

The least common multiple of 2 and 3 is 6.

Multiply the first equation by 6:

3x + 2y = 30

The least common multiple of 4 and 6 is 12.

Multiply the second equation by 12:

3x − 2y = 12

Step 2: Add

6x = 42

x = 7

Step 3: Find y

3(7) + 2y = 30

21 + 2y = 30

2y = 9

y = 9/2 = 4.5

Step 4: Verify

First equation:

7/2 + 4.5/3 = 3.5 + 1.5 = 5

Second equation:

7/4 − 4.5/6 = 1.75 − 0.75 = 1

Answer: x = 7, y = 4.5

Strategy lesson

When a system contains fractions, multiplying each equation by its least common denominator can simplify the calculations significantly.

Recognising the Number of Solutions

A system does not always have exactly one solution.

Unique solution

Consider:

x + y = 8

x − y = 2

Adding:

2x = 10

x = 5

y = 3

The system has one solution.

No solution

Consider:

2x + 4y = 12

x + 2y = 9

Multiplying the second equation by 2 gives:

2x + 4y = 18

But the first equation requires:

2x + 4y = 12

The equations contradict one another.

Conclusion: No solution.

Infinitely many solutions

Consider:

2x + 4y = 12

x + 2y = 6

Multiplying the second equation by 2 produces the first equation.

The equations represent the same line.

Conclusion: Infinitely many solutions.

Geometric interpretation

Integrated Problem: Three-Variable Systems

Some problems involve three unknown quantities and require several stages of elimination.

Solve:

x + y + z = 12

2x + 3y + z = 23

3x + y + 2z = 21

Step 1: Eliminate z

Subtract the first equation from the second:

x + 2y = 11

Multiply the first equation by 2:

2x + 2y + 2z = 24

Subtract this from the third equation:

x − y = −3

The reduced system is:

x + 2y = 11

x − y = −3

Step 2: Solve the reduced system

Subtract the second equation from the first:

3y = 14

y = 14/3

Substitute:

x − 14/3 = −3

x = 5/3

Step 3: Find z

x + y + z = 12

5/3 + 14/3 + z = 12

19/3 + z = 36/3

z = 17/3

Step 4: Verify

First equation:

5/3 + 14/3 + 17/3 = 36/3 = 12

Second equation:

2(5/3) + 3(14/3) + 17/3

= 10/3 + 14 + 17/3

= 27/3 + 14 = 23

Third equation:

3(5/3) + 14/3 + 2(17/3)

= 5 + 48/3 = 21

Answer: x = 5/3, y = 14/3, z = 17/3.

Strategy lesson

Three-variable systems can often be reduced to two-variable systems.

The most important step is eliminating the same variable from two different pairs of equations.

Integrated Problem: Nonlinear Systems

Solve:

x² + y² = 25

y = x + 1

This system contains a circle and a straight line.

Step 1: Choose a method

Substitution is appropriate because y is already isolated.

Step 2: Substitute

x² + (x + 1)² = 25

Expand:

x² + x² + 2x + 1 = 25

2x² + 2x − 24 = 0

Divide by 2:

x² + x − 12 = 0

Step 3: Factor

(x + 4)(x − 3) = 0

Therefore:

x = −4 or x = 3

Step 4: Find y-values

If x = −4:

y = −3

If x = 3:

y = 4

Step 5: Verify

For (−4, −3):

16 + 9 = 25

−3 = −4 + 1

For (3, 4):

9 + 16 = 25

4 = 3 + 1

Answers: (−4, −3) and (3, 4).

Interpretation

The line intersects the circle at two points.

Both solutions are necessary because the original problem asks for all intersections.

Integrated Problem: Business and Economics

A company manufactures two products, A and B.

Each unit of Product A requires 3 hours of labour and 2 kg of material.

Each unit of Product B requires 2 hours of labour and 4 kg of material.

The company uses 160 labour-hours and 160 kg of material.

Determine the number of units manufactured.

Step 1: Define variables

x = number of units of Product A

y = number of units of Product B

Step 2: Construct equations

Labour:

3x + 2y = 160

Materials:

2x + 4y = 160

Step 3: Select a method

Elimination is efficient because the second equation can be divided by 2.

x + 2y = 80

Subtract this from the labour equation:

2x = 80

x = 40

Step 4: Find y

40 + 2y = 80

2y = 40

y = 20

Step 5: Verify

Labour:

3(40) + 2(20) = 160 hours

Materials:

2(40) + 4(20) = 160 kg

Answer: 40 units of Product A and 20 units of Product B.

Evaluate the model

The result is reasonable because both quantities are nonnegative whole numbers.

However, the model assumes constant resource requirements, no waste, and that all recorded resources were used to manufacture these products.

Integrated Problem: Geometry and Algebra

A rectangular garden has a perimeter of 50 m and an area of 150 m².

Find its dimensions.

Step 1: Define variables

l = length, in metres

w = width, in metres

Step 2: Construct equations

Perimeter:

2l + 2w = 50

l + w = 25

Area:

lw = 150

Step 3: Choose a method

Substitution is appropriate because the perimeter equation can easily be rearranged.

l = 25 − w

Step 4: Substitute

(25 − w)w = 150

25w − w² = 150

w² − 25w + 150 = 0

Factor:

(w − 10)(w − 15) = 0

w = 10 or w = 15

The two values represent the same rectangle with the dimensions interchanged.

Answer: The garden measures 15 m by 10 m.

Step 5: Verify

Perimeter:

2(15) + 2(10) = 50 m

Area:

15(10) = 150 m²

Why this problem is important

It requires recognising that one geometric relationship is linear and another is nonlinear.

The problem cannot be solved using ordinary linear elimination alone without additional algebraic manipulation.

Integrated Problem: Science and Experimental Data

A scientist models the position of an object using:

s = vt + s₀

where:

• s = position, in metres.

• v = constant velocity, in m/s.

• t = time, in seconds.

• s₀ = initial position, in metres.

Measurements show:

Determine the velocity and initial position.

Step 1: Construct the equations

11 = 2v + s₀

27 = 6v + s₀

Step 2: Select a method

Elimination is efficient because both equations contain the same coefficient for s₀.

Subtract:

16 = 4v

v = 4 m/s

Step 3: Find initial position

11 = 2(4) + s₀

s₀ = 3 m

Step 4: Write the model

s = 4t + 3

Step 5: Verify

At t = 2 s:

s = 4(2) + 3 = 11 m

At t = 6 s:

s = 4(6) + 3 = 27 m

Answer: Velocity = 4 m/s and initial position = 3 m.

Evaluate

The model assumes constant velocity.

Two measurements determine the parameters of a linear model, but additional measurements are needed to test whether the motion is actually uniform.

Integrated Problem: An Unfamiliar Situation

A wildlife researcher counts two species of animals in a habitat.

Species A has 4 legs per animal.

Species B has 2 legs per animal.

There are 75 animals and 220 legs in total.

Find the number of animals of each species.

Step 1: Understand the situation

The problem does not initially provide algebraic equations.

We must identify the unknown quantities and translate the information into a mathematical model.

Step 2: Define variables

x = number of Species A animals

y = number of Species B animals

Step 3: Construct equations

Total animals:

x + y = 75

Total legs:

4x + 2y = 220

Step 4: Solve

Multiply the first equation by 2:

2x + 2y = 150

Subtract:

2x = 70

x = 35

Then:

y = 75 − 35

y = 40

Step 5: Verify

Total animals:

35 + 40 = 75

Total legs:

4(35) + 2(40)

= 140 + 80

= 220

Answer: 35 animals of Species A and 40 animals of Species B.

Step 6: Evaluate

The model assumes that the habitat contains only these two species and that every animal has the expected number of legs.

In an actual field investigation, sampling error or incomplete observations may affect the results.

General lesson

Unfamiliar problems become easier when we focus on identifying quantities and relationships rather than trying to memorise a particular question type.

Justifying Mathematical Reasoning

A mathematical justification explains why a particular step or method is valid.

Compare these two solutions.

Incomplete explanation

x = 5

y = 3

Complete explanation

The system is:

x + y = 8

x − y = 2

Elimination is efficient because the y-coefficients are opposites.

Adding the equations eliminates y:

2x = 10

Therefore:

x = 5

Substituting into x + y = 8:

y = 3

Checking:

5 + 3 = 8

5 − 3 = 2

Both equations are satisfied, so the solution is (5, 3).

The complete explanation identifies the method, justifies its selection, shows the calculations, and verifies the result.

What makes a strong justification?

A strong mathematical explanation should demonstrate:

• Why the equations represent the problem.

• Why the chosen method is suitable.

• Why each algebraic transformation is valid.

• How the solution satisfies the original conditions.

• What the answer means in context.

Communicating Organised Solutions

A useful structure for written solutions is:

Understand → Define → Model → Solve → Verify → Interpret

Understand

Identify the known information and what must be determined.

Define

Introduce variables with clear meanings and units.

Model

Write the equations representing the relationships.

Solve

Choose an efficient method and show the algebraic steps.

Verify

Substitute the values into the original equations.

Interpret

Explain the answer in the original context.

For real-world problems, add an evaluation of the assumptions and limitations.

This structure makes solutions easier to follow, check, and communicate.

Common Errors in Integrated Problems

Error 1: Choosing an unnecessarily complicated method

Consider:

y = 2x + 3

x + y = 18

Using elimination without rearranging is possible, but substitution is more direct because y is already isolated.

Error 2: Incorrect distribution

Incorrect:

3(x + 2) = 3x + 2

Correct:

3(x + 2) = 3x + 6

Error 3: Changing only one term when multiplying an equation

Incorrect:

2x + y = 10

Multiply by 3:

6x + y = 30

Correct:

6x + 3y = 30

Every term must be multiplied.

Error 4: Losing a negative sign

Incorrect:

−2y = 8

y = 4

Correct:

y = −4

Error 5: Ignoring a second solution

A quadratic equation may produce two valid solutions.

For example:

x² = 16

x = −4 or x = 4

Both must be considered unless the original problem imposes restrictions.

Error 6: Confusing no solution with infinitely many solutions

A contradiction such as:

0 = 5

means no solution.

An identity such as:

0 = 0

means that the eliminated equation provides no additional restriction. In a two-variable linear system where both equations are equivalent, this leads to infinitely many solutions.

Error 7: Failing to interpret units

If x represents time in hours, the answer x = 4 should be communicated as 4 hours.

Error 8: Accepting an unrealistic result

A mathematical solution may be incompatible with the context.

Negative dimensions or fractional counts of complete objects should be investigated rather than automatically accepted.

Strategies for Unfamiliar Problems

When a problem appears unfamiliar, use the following reasoning process.

Identify quantities rather than keywords. Determine what is known and unknown.

Look for conservation relationships. Totals involving mass, cost, distance, or quantity often provide equations.

Look for comparisons. Statements such as “twice as many,” “five greater,” or “20% less” can be translated into equations.

Use a table when appropriate. Organising information often makes the relationships clearer.

Check the number of independent relationships. A unique solution generally requires enough independent information to determine the unknowns.

Choose a method based on the equation structure. Do not automatically use the same method for every problem.

Verify before interpreting. An algebraic error can lead to an incorrect real-world conclusion.

Evaluate the answer. Ask whether the values make sense and satisfy any practical restrictions.

Key Terms

• System of equations: Two or more equations solved simultaneously.

• Simultaneous solution: Values satisfying every equation in a system.

• Substitution: Replacing a variable with an equivalent expression.

• Elimination: Combining equations to remove a variable.

• Graphical method: Solving or representing a system using graphs.

• Matrix method: Representing and solving a system using matrices.

• Linear system: A system consisting of linear equations.

• Nonlinear system: A system containing at least one nonlinear equation.

• Unique solution: Exactly one solution satisfying the system.

• Inconsistent system: A system with no solution.

• Dependent equations: Equations that do not provide independent information.

• Equivalent equations: Equations having exactly the same solution set.

• Mathematical model: A mathematical representation of a real-world situation.

• Constraint: A restriction on possible solutions.

• Verification: Checking whether a solution satisfies the original equations.

• Interpretation: Explaining the mathematical result in context.

• Justification: Explaining why a mathematical method or step is valid.

• Efficiency: Solving a problem with a suitable balance of simplicity, accuracy, and clarity.

• Extraneous solution: A candidate solution that does not satisfy the original problem.

• Independent information: Information that contributes a new mathematical restriction.

Key Takeaways

• Different systems may be best solved using different methods.

• Substitution is often efficient when a variable is isolated.

• Elimination is often efficient when coefficients can cancel.

• Graphing provides a visual interpretation of solutions.

• Matrix methods and technology are useful for larger systems.

• Three-variable systems can often be reduced to two-variable systems.

• Nonlinear systems may produce multiple solutions.

• Some systems have no solution or infinitely many solutions.

• Real-world problems require variables, equations, constraints, and interpretation.

• Mathematical reasoning should be justified and clearly communicated.

• Every solution should be checked against the original equations.

• Unfamiliar problems can be approached systematically by identifying quantities and relationships.

Check Your Understanding

For each problem, select an efficient method, show your working, and verify your solution. For application questions, explain what your answer means in context.

Part A: Selecting Methods

1. Identify the most efficient method for solving each system and explain your choice.

a) y = 4x − 3 and x + y = 17

b) 3x + 2y = 16 and 5x − 2y = 8

c) y = x² − 4 and y = 2x − 4

d) x + y + z = 12, 2x − y + z = 9, and x + 2y − z = 3

2. Explain why elimination is usually efficient when two equations contain opposite coefficients for one variable.

3. Explain why graphing may not always provide an exact solution.

Part B: Algebraic Systems

4. Solve:

2x + 3y = 19

4x − 3y = 5

5. Solve:

y = 3x − 2

2x + y = 18

6. Solve:

x/2 + y/4 = 5

x/3 − y/6 = 1

7. Determine whether the following system has one solution, no solution, or infinitely many solutions:

3x + 6y = 18

x + 2y = 6

8. Determine the number of solutions:

2x − 4y = 10

x − 2y = 8

Part C: Advanced Systems

9. Solve:

x + y + z = 10

2x − y + z = 9

x + 2y − z = 3

10. Find all intersection points of:

x² + y² = 25

y = x + 1

11. Solve:

xy = 24

x + y = 10

Part D: Integrated Applications

12. A school sells adult tickets for $18 and student tickets for $12.

A total of 150 tickets are sold, generating $2,250.

Determine the number of each ticket type.

13. A rectangle has a perimeter of 46 cm and an area of 120 cm².

Find its dimensions.

14. Two vehicles travel in opposite directions from the same starting point.

After 3 hours, they are 270 km apart.

One vehicle travels 10 km/h faster than the other.

Find their speeds.

15. A laboratory needs 400 mL of a 30% solution.

It has a 20% solution and a 50% solution.

Assuming compatible concentrations and additive volumes, determine how much of each is required.

16. A company manufactures two products.

Product A requires 4 labour-hours and 3 units of material.

Product B requires 2 labour-hours and 5 units of material.

The company uses 100 labour-hours and 110 units of material.

Determine the production quantities and evaluate whether the result is realistic.

Part E: Analysis and Mathematical Reasoning

17. A student solves:

2x + y = 12

x − y = 3

and obtains x = 3, y = 6.

Identify the error by checking the solution, then find the correct answer.

18. A system produces the solution:

x = −5

y = 12

Explain why this may be acceptable in coordinate geometry but unacceptable in a problem involving physical dimensions.

19. A scientist constructs a two-variable model using two experimental measurements. The model has a unique solution, but a third measurement does not agree with its predictions.

Explain what the scientist should investigate.

20. Integrated Challenge

A school sells three types of merchandise:

• Pens for $2 each.

• Notebooks for $5 each.

• T-shirts for $12 each.

A total of 100 items are sold for $500.

The number of pens sold is twice the number of T-shirts.

Determine the number of each item, verify your answer, and explain why a three-variable system is appropriate.

Answers to Check Your Understanding

1. Selecting methods

a) Substitution. The first equation already isolates y.

b) Elimination. The y-coefficients are +2 and −2, so they cancel immediately.

c) Substitution. Both equations express y in terms of x, allowing the expressions to be set equal.

d) Elimination or matrices. Three unknowns require combining equations systematically.

2. Why opposite coefficients help

When the equations are added, the terms with opposite coefficients cancel, leaving an equation with fewer unknowns.

This reduces the amount of algebra required.

3. Graphing limitations

A graph may show only approximate intersection coordinates.

The graphing window, display resolution, and scale can also affect accuracy.

Algebraic methods are often preferable when exact values are required.

4. Elimination

2x + 3y = 19

4x − 3y = 5

Add:

6x = 24

x = 4

Substitute:

2(4) + 3y = 19

3y = 11

y = 11/3

Answer: (4, 11/3).

Verification:

2(4) + 3(11/3) = 19

4(4) − 3(11/3) = 5

5. Substitution

y = 3x − 2

2x + y = 18

Substitute:

2x + 3x − 2 = 18

5x = 20

x = 4

y = 3(4) − 2 = 10

Answer: (4, 10).

6. Fractions

x/2 + y/4 = 5

Multiply by 4:

2x + y = 20

x/3 − y/6 = 1

Multiply by 6:

2x − y = 6

Add:

4x = 26

x = 13/2

Then:

y = 20 − 2(13/2)

y = 7

Answer: x = 13/2, y = 7.

Verification:

(13/2)/2 + 7/4 = 13/4 + 7/4 = 5

(13/2)/3 − 7/6 = 13/6 − 7/6 = 1

7. Infinitely many solutions

3x + 6y = 18

x + 2y = 6

Multiplying the second equation by 3 produces the first.

Answer: Infinitely many solutions.

The equations represent the same line.

8. No solution

2x − 4y = 10

x − 2y = 8

Multiply the second equation by 2:

2x − 4y = 16

The same expression cannot equal both 10 and 16.

Answer: No solution.

9. Three-variable system

x + y + z = 10

2x − y + z = 9

x + 2y − z = 3

Subtract the first equation from the second:

x − 2y = −1

Add the first and third equations:

2x + 3y = 13

From the first reduced equation:

x = 2y − 1

Substitute:

2(2y − 1) + 3y = 13

7y − 2 = 13

7y = 15

y = 15/7

Then:

x = 2(15/7) − 1 = 23/7

Using the first original equation:

z = 10 − 23/7 − 15/7

z = 32/7

Answer: x = 23/7, y = 15/7, z = 32/7.

Verification:

First equation: (23 + 15 + 32)/7 = 10

Second equation: (46 − 15 + 32)/7 = 9

Third equation: (23 + 30 − 32)/7 = 3

10. Line-circle intersections

Substitute y = x + 1:

x² + (x + 1)² = 25

2x² + 2x − 24 = 0

x² + x − 12 = 0

(x + 4)(x − 3) = 0

x = −4 or x = 3

Corresponding y-values:

y = −3 or y = 4

Answer: (−4, −3) and (3, 4).

11. Product and sum

xy = 24

x + y = 10

Rearrange:

y = 10 − x

Substitute:

x(10 − x) = 24

10x − x² = 24

x² − 10x + 24 = 0

(x − 4)(x − 6) = 0

x = 4 or x = 6

Answers: (4, 6) and (6, 4).

Both satisfy the original system.

12. Ticket sales

Let:

a = adult tickets

s = student tickets

a + s = 150

18a + 12s = 2250

Multiply the first equation by 12:

12a + 12s = 1800

Subtract:

6a = 450

a = 75

Then:

s = 75

Answer: 75 adult tickets and 75 student tickets.

Verification:

18(75) + 12(75) = $2250

13. Rectangle dimensions

Let:

l = length

w = width

Perimeter:

l + w = 23

Area:

lw = 120

Substitute l = 23 − w:

(23 − w)w = 120

w² − 23w + 120 = 0

(w − 8)(w − 15) = 0

Answer: 15 cm by 8 cm.

Verification:

2(15) + 2(8) = 46 cm

15(8) = 120 cm²

14. Vehicle speeds

Let:

x = slower speed

y = faster speed

3x + 3y = 270

x + y = 90

y − x = 10

Add:

2y = 100

y = 50

x = 40

Answer: 40 km/h and 50 km/h.

Verification:

3(40) + 3(50) = 270 km

15. Solution mixture

Let:

x = volume of 20% solution

y = volume of 50% solution

x + y = 400

0.20x + 0.50y = 120

Substitute x = 400 − y:

80 − 0.20y + 0.50y = 120

0.30y = 40

y = 400/3 ≈ 133.33 mL

Then:

x = 800/3 ≈ 266.67 mL

Answer: Approximately 266.67 mL of 20% solution and 133.33 mL of 50% solution.

Use the exact fractional values for verification to avoid rounding error.

16. Manufacturing

Let:

x = Product A units

y = Product B units

Labour:

4x + 2y = 100

Materials:

3x + 5y = 110

Divide the labour equation by 2:

2x + y = 50

Therefore:

y = 50 − 2x

Substitute:

3x + 5(50 − 2x) = 110

3x + 250 − 10x = 110

−7x = −140

x = 20

Then:

y = 10

Answer: 20 units of Product A and 10 units of Product B.

Verification:

4(20) + 2(10) = 100 hours

3(20) + 5(10) = 110 units of material

The quantities are nonnegative whole numbers, so they are feasible under the stated assumptions.

17. Error analysis

The student's proposed solution is:

x = 3, y = 6

Check the first equation:

2(3) + 6 = 12

This is correct.

Check the second equation:

3 − 6 = −3

But the equation requires +3.

Therefore, the proposed values do not solve the system.

To correct the solution:

2x + y = 12

x − y = 3

Add:

3x = 15

x = 5

Then:

5 − y = 3

y = 2

Correct answer: (5, 2).

Both equations are satisfied.

18. Negative solutions

Negative coordinates are valid in coordinate geometry.

For example, x = −5 may represent a point 5 units to the left of the y-axis.

However, a negative physical length or width is not meaningful in ordinary geometric measurement.

The context determines whether a negative value is acceptable.

19. Model evaluation

The scientist should investigate whether the model's assumptions are appropriate.

Possible issues include measurement uncertainty, incorrect data, changing experimental conditions, or a relationship that is not truly linear.

Additional measurements and a revised model may be necessary.

20. Integrated challenge

Let:

p = number of pens

n = number of notebooks

t = number of T-shirts

Total items:

p + n + t = 100

Revenue:

2p + 5n + 12t = 500

Quantity relationship:

p = 2t

Substitute into the first equation:

2t + n + t = 100

n = 100 − 3t

Substitute into the revenue equation:

2(2t) + 5(100 − 3t) + 12t = 500

4t + 500 − 15t + 12t = 500

t = 0

Then:

p = 0

n = 100

Answer: 0 pens, 100 notebooks, and 0 T-shirts.

Verification:

Total items:

0 + 100 + 0 = 100

Revenue:

2(0) + 5(100) + 12(0) = $500

Quantity relationship:

0 = 2(0)

All conditions are satisfied.

The system requires three variables because the school sells three different types of merchandise.

The result is mathematically valid, although it is a boundary case in which only notebooks were sold. If the problem required at least one of every item, the information would be inconsistent with that additional requirement.

Final Summary

Solving systems of equations confidently requires more than memorising algebraic procedures.

It involves recognising mathematical structures, choosing efficient methods, constructing models, justifying calculations, and interpreting solutions.

The strongest problem-solvers are flexible. They can use substitution when a variable is isolated, elimination when coefficients can cancel, graphing when visual interpretation is useful, and technology when systems become more complex.

They also recognise when a system has no solution, infinitely many solutions, or multiple solutions arising from nonlinear relationships.

Most importantly, they verify their calculations and communicate their conclusions clearly.

Successful integrated problem-solving means choosing an appropriate strategy, applying it accurately, explaining why it works, and determining whether the result makes sense.