Algebraic Problem Solving

Introduction

Algebra becomes especially useful when we apply it to problems outside the mathematics classroom. Many everyday situations involve unknown quantities, relationships between values, and conditions that can be represented using equations.

For example, suppose a school is organising a field trip. The bus company charges a fixed fee of $150 plus $12 for each student. The total transportation cost is $510.

How many students are attending?

We can represent the situation using an equation.

Let n = number of students.

Total cost = fixed fee + cost per student

510 = 150 + 12n

Subtract 150 from both sides:

360 = 12n

Divide by 12:

n = 30

Therefore, 30 students are attending the trip.

We can verify the answer:

150 + 12(30) = 510

150 + 360 = 510 ✓

This example demonstrates how a real-world situation can be translated into mathematics, solved using algebra, and interpreted in context.

In this topic, we will explore how to construct equations from written descriptions, solve practical algebra problems, evaluate the reasonableness of solutions, and communicate mathematical reasoning clearly.

What Are Applied Equations?

An applied equation is an equation used to represent and solve a problem involving a real-world situation.

Applied equations are common in:

• Personal finance and budgeting.

• Shopping and pricing.

• Travel, speed, distance, and time.

• Science and engineering.

• Geometry and measurement.

• Business and profit calculations.

• Mixtures and concentration.

• Population growth.

• Planning and resource allocation.

The algebraic techniques are often familiar, but the challenge is determining which equation represents the situation correctly.

A useful problem-solving process is:

Understand → Define → Model → Solve → Check → Interpret

Understanding the Problem Before Solving

Before writing an equation, identify three important elements.

Known information

What values or relationships are provided?

For example:

A concert ticket costs $18, and there is a fixed booking fee of $6.

Known information:

• Ticket price = $18.

• Booking fee = $6.

Unknown information

What are we trying to calculate?

For example:

How many tickets can be purchased for $150?

Unknown:

Number of tickets.

Relationships

How are the quantities connected?

Total cost = ticket cost + booking fee

If n is the number of tickets:

C = 18n + 6

If the total cost is exactly $150:

150 = 18n + 6

Solve:

144 = 18n

n = 8

Therefore, eight tickets can be purchased for exactly $150.

If the question instead asks for the maximum number of tickets affordable with a budget of $150, the mathematical model is an inequality:

18n + 6 ≤ 150

This also gives a maximum of eight tickets, but the distinction between an exact total and a maximum budget is important.

Defining Variables Clearly

A variable represents an unknown or changing quantity.

Before constructing an equation, clearly state what each variable represents.

For example:

Let x = number of adult tickets sold.

This is better than simply writing "Let x = tickets" because it identifies the quantity precisely.

Variables may represent:

• Whole numbers, such as the number of students.

• Measurements, such as length or mass.

• Monetary values.

• Time intervals.

• Rates, such as speed or hourly wages.

• Quantities that may be positive or negative.

Restrictions on variables

Some variables have natural restrictions.

For example:

• The number of students must be a non-negative integer.

• A physical length cannot be negative.

• A time interval cannot be negative.

• A percentage discount is usually between 0% and 100%.

• A denominator cannot equal zero.

Recognising these restrictions helps us determine whether a mathematical solution is reasonable.

Translating Words into Equations

Many written statements can be translated directly into algebra.

Worked Example: Translating a statement

Three times a number, increased by eight, equals 50.

Let x = the unknown number.

Translate:

3x + 8 = 50

Subtract 8:

3x = 42

Divide by 3:

x = 14

Verification:

3(14) + 8 = 50

The number is 14.

Worked Example: Consecutive integers

The sum of three consecutive integers is 84.

Let n = the smallest integer.

The integers are:

n, n + 1, n + 2

Construct the equation:

n + (n + 1) + (n + 2) = 84

Simplify:

3n + 3 = 84

Subtract 3:

3n = 81

Divide by 3:

n = 27

The integers are:

27, 28, and 29

Verification:

27 + 28 + 29 = 84 ✓

Solving Problems Involving Money

Financial situations are particularly suitable for algebraic equations because they often involve fixed costs, rates, and totals.

Worked Example: Shopping

A student purchases several notebooks costing $4 each and a calculator costing $28.

The total cost is $60.

How many notebooks were purchased?

Let n = number of notebooks.

Total cost:

4n + 28 = 60

Subtract 28:

4n = 32

Divide by 4:

n = 8

Answer: Eight notebooks were purchased.

Verification:

8(4) + 28 = 32 + 28 = $60

Worked Example: Saving money

A student already has $85 saved and adds $15 each week.

How many weeks will it take to have $250?

Let w = number of weeks.

Total savings:

S = 85 + 15w

Set the total to $250:

250 = 85 + 15w

Subtract 85:

165 = 15w

Divide by 15:

w = 11

Answer: The student will reach $250 after 11 weeks.

Solving Problems Involving Distance, Speed, and Time

One of the most common scientific relationships is:

d = vt

where:

• d = distance travelled.

• v = constant speed.

• t = time taken.

The formula can be rearranged:

v = d/t

t = d/v

The correct rearrangement depends on the unknown quantity.

Worked Example: Finding travel time

A train travels at a constant speed of 90 km/h.

How long does it take to travel 225 km?

Given:

d = 225 km

v = 90 km/h

Use:

t = d/v

Substitute:

t = 225/90

t = 2.5 hours

Answer: The journey takes 2 hours and 30 minutes.

Worked Example: Two vehicles travelling toward each other

Two towns are 300 km apart.

A car leaves one town travelling at 80 km/h. At the same time, another car leaves the other town travelling toward the first at 70 km/h.

How long will it take for the cars to meet?

Let t = time in hours.

Distance travelled by the first car:

d₁ = 80t

Distance travelled by the second car:

d₂ = 70t

Together, they travel 300 km:

80t + 70t = 300

150t = 300

t = 2

Answer: The cars meet after 2 hours.

Verification:

First car: 80 × 2 = 160 km

Second car: 70 × 2 = 140 km

Total = 300 km ✓

Solving Problems Involving Geometry

Geometric problems often require us to combine algebraic expressions with formulas for perimeter, area, and volume.

Worked Example: Rectangle dimensions

A rectangle has a perimeter of 64 cm.

Its length is 8 cm greater than its width.

Find its dimensions.

Let w = width in centimetres.

Length:

l = w + 8

Perimeter:

P = 2l + 2w

Substitute:

64 = 2(w + 8) + 2w

Expand:

64 = 2w + 16 + 2w

Simplify:

64 = 4w + 16

Subtract 16:

48 = 4w

w = 12

Length:

l = 12 + 8 = 20

Answer: Width = 12 cm, length = 20 cm.

Verification:

P = 2(20) + 2(12)

P = 40 + 24

P = 64 cm ✓

Worked Example: Area of a rectangle

A rectangular garden has an area of 96 m². Its length is 4 m greater than its width.

Find its dimensions.

Let w = width in metres.

Length = w + 4

Area:

A = lw

Substitute:

96 = w(w + 4)

Expand:

96 = w² + 4w

Rearrange:

w² + 4w − 96 = 0

Factor:

(w + 12)(w − 8) = 0

Possible solutions:

w = −12 or w = 8

A negative width is not physically meaningful.

Therefore:

w = 8 m

Length = 8 + 4 = 12 m

Answer: The garden measures 8 m by 12 m.

Verification:

8 × 12 = 96 m² ✓

Solving Problems Involving Ages

Age problems often involve relationships between people's ages at different times.

The important principle is that everyone ages at the same rate: after five years, each person's age increases by five years.

Worked Example: Present ages

A father is three times as old as his son.

Their combined age is 56 years.

Find their present ages.

Let x = son's age.

Father's age = 3x

Combined age:

x + 3x = 56

4x = 56

x = 14

Father's age:

3(14) = 42

Answer: The son is 14 years old, and the father is 42 years old.

Verification:

14 + 42 = 56

42 = 3 × 14

Both conditions are satisfied.

Worked Example: Ages in the future

A mother is currently 36 years old, and her daughter is 12.

In how many years will the mother be twice as old as her daughter?

Let t = number of years from now.

Mother's future age:

36 + t

Daughter's future age:

12 + t

Construct the equation:

36 + t = 2(12 + t)

Expand:

36 + t = 24 + 2t

Subtract 24:

12 + t = 2t

Subtract t:

t = 12

Answer: In 12 years, the mother will be twice as old as her daughter.

Verification:

Mother: 36 + 12 = 48

Daughter: 12 + 12 = 24

48 = 2 × 24 ✓

Solving Problems Involving Mixtures

Mixture problems often involve combining substances with different concentrations or prices.

The central idea is that the total amount of a substance is conserved when the mixtures are combined, assuming no substance is lost or produced.

Worked Example: Mixing solutions

A chemist needs 500 mL of a 20% salt solution.

Two solutions are available:

• A 10% salt solution.

• A 30% salt solution.

How much of each should be mixed?

Let x = volume of 10% solution in mL.

Volume of 30% solution:

500 − x

The amount of salt in the final mixture must equal the sum of the amounts of salt in the original solutions.

Construct the equation:

0.10x + 0.30(500 − x) = 0.20(500)

Expand:

0.10x + 150 − 0.30x = 100

Simplify:

−0.20x + 150 = 100

Subtract 150:

−0.20x = −50

Divide:

x = 250

Therefore:

10% solution = 250 mL

30% solution = 500 − 250 = 250 mL

Answer: Mix 250 mL of each solution.

Verification:

Salt from first solution:

0.10 × 250 = 25 mL equivalent

Salt from second solution:

0.30 × 250 = 75 mL equivalent

Total = 100 mL equivalent

100/500 = 0.20 = 20%

The model assumes the percentages are volume-based and that the volumes are additive.

Solving Problems Involving Percentages

Percentages appear in discounts, taxes, interest, population changes, and many other applications.

Worked Example: Finding an original price

A jacket is discounted by 25% and now costs $90.

What was the original price?

Let P = original price.

After a 25% discount, the customer pays 75% of the original price.

0.75P = 90

Divide by 0.75:

P = 90/0.75

P = 120

Answer: The original price was $120.

Verification:

25% of $120 = $30

$120 − $30 = $90 ✓

Worked Example: Percentage increase

A town's population increases by 8% and becomes 27,000.

What was the original population?

Let P = original population.

After an 8% increase:

1.08P = 27,000

Divide:

P = 27,000/1.08

P = 25,000

Answer: The original population was 25,000.

Verification:

8% of 25,000 = 2,000

25,000 + 2,000 = 27,000 ✓

Solving Problems with Systems of Equations

Sometimes a real-world situation involves two unknown quantities.

In such cases, we may need two equations.

Worked Example: Ticket sales

A school sells adult tickets for $15 and student tickets for $8.

A total of 100 tickets are sold, producing $1,080.

How many of each ticket were sold?

Let:

a = number of adult tickets

s = number of student tickets

Total tickets:

a + s = 100

Total revenue:

15a + 8s = 1080

Rearrange the first equation:

s = 100 − a

Substitute into the second:

15a + 8(100 − a) = 1080

Expand:

15a + 800 − 8a = 1080

Simplify:

7a + 800 = 1080

Subtract 800:

7a = 280

a = 40

Calculate student tickets:

s = 100 − 40

s = 60

Answer: 40 adult tickets and 60 student tickets.

Verification:

40 + 60 = 100 tickets

15(40) + 8(60) = 600 + 480 = $1,080

Both conditions are satisfied.

Solving Applied Quadratic Equations

Some real-world problems produce quadratic equations.

These often arise when calculating areas, dimensions, and motion under constant acceleration.

Worked Example: Projectile motion

A ball is thrown upward from a platform. Its height above the ground is modeled by:

h = −5t² + 20t + 15

where:

• h = height above ground in metres.

• t = time in seconds.

The model approximates gravitational acceleration as 10 m/s² and neglects air resistance.

When does the ball reach the ground?

At ground level:

h = 0

Therefore:

−5t² + 20t + 15 = 0

Divide by −5:

t² − 4t − 3 = 0

Use the quadratic formula:

t = (4 ± √(16 + 12))/2

t = (4 ± √28)/2

t = 2 ± √7

Approximate solutions:

t ≈ 4.646 or t ≈ −0.646

The negative time is outside the physical domain of this situation.

Answer: The ball reaches the ground approximately 4.65 seconds after being thrown.

Evaluating Whether Solutions Are Reasonable

Solving an equation is only part of the process.

We must also determine whether the result makes sense.

Checking units

If a question asks for distance, the answer should use a distance unit such as metres or kilometres.

Checking signs

Negative values may be valid for quantities such as displacement, temperature in Celsius, or financial losses.

However, negative values are generally invalid for physical lengths, elapsed time, and numbers of people.

Checking whole-number restrictions

If a calculation gives:

n = 12.7 buses

we cannot use 12.7 actual buses.

If the problem asks for the minimum number needed, we would normally round up to 13 buses.

Checking magnitude

Suppose a student calculates that a cyclist travels 5,000 km in 10 minutes.

Although the algebra may appear correct, the result is not reasonable for an ordinary cyclist.

This suggests a possible error in units, substitution, or calculation.

Checking original conditions

A solution must satisfy every condition in the problem, not just one equation.

For example, a ticket problem may require both the correct total number of tickets and the correct total revenue.

Using Estimation to Check Answers

Estimation can help identify unreasonable results before performing exact calculations.

Worked Example: Estimating a journey time

A car travels 310 km at approximately 100 km/h.

Estimated time:

t ≈ 310/100

t ≈ 3.1 hours

If a detailed calculation gives 31 hours, something has gone wrong.

Estimation provides a useful independent check.

Worked Example: Checking a shopping calculation

A student buys 19 items costing approximately $5 each.

Estimated total:

19 × 5 ≈ $95

A calculated total of $950 is unlikely to be correct.

Estimation does not replace exact calculations, but it can help detect mistakes.

Communicating Solutions Clearly

A well-presented applied algebra solution should include:

1. Define the variable: Explain what the unknown represents.

2. Construct the equation: Translate the situation into mathematics.

3. Solve: Show each algebraic operation logically.

4. Verify: Substitute the result into the original relationship.

5. Interpret: State the answer using appropriate units and context.

Complete Worked Example: Planning a School Event

A school is organising a science fair.

The venue costs $350 to rent, and materials cost $8 per student.

The total available budget is $950.

How many students can participate?

Step 1: Define the variable

Let n = number of students.

Step 2: Construct the model

Total cost:

C = 350 + 8n

The cost must not exceed $950:

350 + 8n ≤ 950

Step 3: Solve

Subtract 350:

8n ≤ 600

Divide by 8:

n ≤ 75

Step 4: Interpret

A maximum of 75 students can participate within the budget.

Step 5: Verify

C = 350 + 8(75)

C = 350 + 600

C = $950

The total exactly matches the budget.

Step 6: Evaluate

The model assumes the cost per student remains constant and that no additional expenses are incurred.

This solution combines algebra, inequality reasoning, practical interpretation, and verification.

Common Mistakes

Failing to identify the unknown

Before solving, clearly define the quantity represented by the variable.

Constructing the wrong equation

Read the problem carefully and identify how the quantities are related. A correct algebraic solution to an incorrect equation will not solve the original problem.

Ignoring fixed costs

Many practical problems involve both a fixed fee and a variable cost.

For example:

C = 50 + 12n

The fixed fee of $50 must be included.

Confusing total and remaining quantities

If 100 tickets are sold and x are adult tickets, the number of student tickets is 100 − x, not x.

Using inconsistent units

Convert measurements into compatible units before substituting them into equations.

Accepting impossible solutions

Negative lengths, fractional people, or times outside the situation's domain may need to be rejected.

Rounding incorrectly

When calculating the minimum number of vehicles required, round up. When calculating the maximum number of whole items affordable, round down.

Failing to verify

Always check that the solution satisfies the original conditions.

Not interpreting the result

An answer such as x = 15 is incomplete if the question asks how many students, hours, or kilometres are involved.

Key Terms

• Applied equation: An equation used to represent and solve a real-world problem.

• Variable: A symbol representing an unknown or changing quantity.

• Constant: A value that remains unchanged within a model.

• Coefficient: A numerical factor multiplying a variable.

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

• Linear equation: An equation involving variables to the first power.

• Quadratic equation: An equation involving a squared variable.

• System of equations: Two or more equations that must be satisfied simultaneously.

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

• Elimination: Removing a variable by combining equations.

• Constraint: A condition restricting possible values.

• Inequality: A statement comparing quantities using symbols such as <, >, ≤, or ≥.

• Reasonableness: Whether a mathematical result makes sense in context.

• Estimation: Finding an approximate value to guide or check calculations.

• Verification: Checking a solution against the original equation or conditions.

• Interpretation: Explaining the meaning of a result in its real-world context.

• Fixed cost: A cost that does not change with the number of items or activities.

• Variable cost: A cost that changes with the quantity produced, purchased, or used.

• Rate: A comparison of two quantities with different units.

• Domain: The set of allowed values for a variable.

• Extraneous solution: A value obtained during solving that does not satisfy the original equation.

Key Takeaways

• Applied equations connect algebra to practical situations.

• Understanding the problem is essential before constructing an equation.

• Variables should be clearly defined and given appropriate units.

• Written relationships can be translated into equations or inequalities.

• Real-world problems may involve linear, quadratic, or simultaneous equations.

• Fixed costs, variable costs, rates, and totals must be identified correctly.

• Mathematical solutions must be checked against the original conditions.

• Not every algebraic solution is physically or practically meaningful.

• Estimation helps identify unreasonable results.

• Clear mathematical communication includes equations, working, units, verification, and interpretation.

Check Your Understanding

Recall and Understanding

1. What is an applied equation?

2. Why is it important to define variables before constructing an equation?

3. Explain the difference between an equation and an inequality in a budgeting problem.

4. Give two reasons why a mathematically correct answer might not be reasonable in a real-world situation.

Application

5. A student purchases notebooks costing $6 each and pays a fixed delivery fee of $9. The total cost is $57. How many notebooks were purchased?

6. A cyclist travels 84 km at a constant speed of 21 km/h. How long does the journey take?

7. A rectangle has a perimeter of 50 cm. Its length is 5 cm greater than its width. Find its dimensions.

8. A jacket is sold for $108 after a 20% discount. What was the original price?

9. A father is four times as old as his daughter. Their combined age is 65 years. Find their ages.

10. A school purchases 50 tickets for a performance. Adult tickets cost $14, and student tickets cost $8. The total cost is $520. How many adult and student tickets were purchased?

Analysis and Challenge

11. Two taxi companies charge different rates.

Company A charges $5 plus $2 per kilometre.

Company B charges $11 plus $1.50 per kilometre.

At what distance will both companies charge the same amount?

12. A rectangular garden has an area of 120 m². Its length is 2 m greater than its width. Calculate its dimensions.

13. A ball's height is modeled by:

h = −5t² + 15t + 20

where h is measured in metres and t in seconds.

Calculate when the ball reaches the ground.

14. A school has a budget of $800 for purchasing science equipment. Each kit costs $35, and delivery costs $65. Determine the maximum number of kits the school can purchase.

15. A student solves a word problem and obtains x = −6, where x represents the number of students attending an event. Explain why the result is not reasonable and describe what the student should do next.

Answers to Check Your Understanding

Final Summary

Solving applied equations involves more than manipulating algebraic symbols. It requires understanding real situations, identifying relationships, constructing appropriate equations, and interpreting results.

A successful solution follows a logical process: understand the situation, define the variables, construct the equation, solve carefully, verify the answer, and explain its meaning.

The true value of applied algebra is its ability to turn real-world questions into mathematical problems that can be solved, checked, and used to make informed decisions.