Algebra in Everyday Life

Algebra is not limited to mathematics classrooms. We use algebra whenever we work with unknown values, changing quantities, patterns, costs, distances, time, measurements, or predictions.

When we use a letter or symbol to represent a quantity, we can create a mathematical model of a real situation. That model can then help us calculate values, compare choices, solve problems, and predict what might happen.

Learning Targets

• I can identify situations that can be represented using algebra.
• I can use variables to model real-world relationships.
• I can create algebraic expressions from practical scenarios.
• I can use algebra to solve everyday problems.
• I can explain how algebra helps make predictions and decisions.

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Where Do We Use Algebra?

Algebra appears whenever one quantity depends on another.

Examples include:

• Calculating the cost of several items
• Planning a journey
• Comparing phone plans
• Creating a budget
• Calculating wages
• Converting temperatures
• Working out discounts
• Calculating areas and volumes
• Predicting future savings
• Comparing different prices
• Planning how much material is needed

In many of these situations, the quantities can change.

Variables allow us to create a rule that works for many different values, rather than completing a separate calculation every time.

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Variables Represent Real Quantities

A variable is a letter or symbol that represents a value.

In real-world problems, variables usually represent quantities with specific meanings.

For example:

t = time

d = distance

s = speed

c = cost

n = number of items

h = number of hours worked

The important idea is that the variable is not simply a random letter.

It represents a real quantity in the situation being modeled.

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Example — Buying Movie Tickets

Suppose one movie ticket costs:

$12

If you buy one ticket:

Cost = $12

If you buy two:

Cost = $24

If you buy five:

Cost = $60

Instead of writing a new calculation every time, let:

n = number of tickets

Then:

Cost = 12n

This algebraic expression works for any number of tickets.

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From a Situation to an Expression

When creating an algebraic expression, ask:

What quantity can change?

Choose a variable for it.

Then ask:

How are the quantities related?

Finally, translate that relationship into mathematics.

For example:

A notebook costs $4.

Let:

n = number of notebooks

Then:

Total cost = 4n

If there is also a $3 delivery fee:

Total cost = 4n + 3

The variable part represents the changing cost.

The constant represents the fixed cost.

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Fixed and Variable Costs

Many real-world situations contain both fixed costs and variable costs.

A fixed cost does not change.

A variable cost depends on how much of something is used or purchased.

Suppose a bicycle rental company charges:

$8 starting fee

plus:

$5 per hour

Let:

h = number of hours

Then:

Cost = 5h + 8

Here:

5h is the variable cost.

8 is the fixed cost.

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Worked Example — Taxi Fare

A taxi company charges:

$4 starting fee

plus:

$2.50 per kilometre

Let:

d = distance travelled in kilometres

The distance cost is:

2.5d

Add the starting fee:

C = 2.5d + 4

where C represents the total cost.

If the journey is 10 km:

C = 2.5(10) + 4

C = 25 + 4

C = $29

Algebra allows the same rule to be used for journeys of any distance.

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Algebra and Shopping

Shopping provides many opportunities to use algebra.

Suppose apples cost:

$3 per kilogram

Let:

m = mass of apples in kilograms

Then:

Cost = 3m

If you purchase 2.5 kg:

Cost = 3(2.5)

Cost = $7.50

This type of algebra is used automatically by supermarket scales and checkout systems.

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Algebra and Discounts

Algebra can also represent discounts.

Suppose an item originally costs:

p dollars

and receives a:

20% discount

The discount is:

0.20p

The new price is:

p − 0.20p

which simplifies to:

0.80p

Therefore:

Sale price = 0.80p

If the original price is $60:

Sale price = 0.80(60)

Sale price = $48

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Worked Example — Sale Price

A jacket costs:

p dollars

A store offers a:

25% discount

After the discount, the customer pays:

75% of the original price

Therefore:

Sale price = 0.75p

If:

p = $120

then:

Sale price = 0.75(120)

Sale price = $90

Algebra makes it easy to apply the same discount rule to any original price.

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Algebra and Budgeting

A budget is essentially a mathematical model of income and expenses.

Suppose you have:

$500

and spend:

$40 each week

Let:

w = number of weeks

Money spent:

40w

Money remaining:

500 − 40w

After 6 weeks:

500 − 40(6)

= 500 − 240

= $260

The expression can also help predict how much money will remain after any number of weeks.

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Algebra and Wages

Suppose a worker earns:

$18 per hour

Let:

h = number of hours worked

Then:

Pay = 18h

If the worker completes 30 hours:

Pay = 18(30)

Pay = $540

If there is also a fixed $50 bonus:

Pay = 18h + 50

This model can be used to predict earnings before the work is completed.

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Algebra and Travel

Travel involves relationships between:

• Distance
• Speed
• Time

One important relationship is:

Distance = Speed × Time

or:

d = vt

where:

• d = distance
• v = speed
• t = time

This is an algebraic model.

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Worked Example — Journey

A car travels at:

80 km/h

Let:

t = travel time in hours

Distance travelled is:

d = 80t

After 2 hours:

d = 80(2)

d = 160 km

After 3.5 hours:

d = 80(3.5)

d = 280 km

The expression allows us to predict distance for any travel time, assuming the speed remains constant.

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Solving an Everyday Problem

Sometimes we know the result and want to find the unknown value.

Suppose a taxi charges:

C = 3d + 5

where:

• C = total cost
• d = distance in kilometres

A journey costs:

$35

How far was the journey?

Substitute:

35 = 3d + 5

Subtract 5:

30 = 3d

Divide by 3:

d = 10

Therefore:

The journey was 10 km.

Algebra has allowed us to work backward from the cost to determine the distance.

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Comparing Two Options

Algebra is particularly useful when making decisions.

Imagine two bicycle rental companies.

Company A charges:

$10 starting fee + $4 per hour

Company B charges:

$6 per hour with no starting fee

Let:

h = number of hours

Company A:

A = 4h + 10

Company B:

B = 6h

These expressions allow us to compare the companies for different rental times.

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Worked Example — Comparing Plans

For a 2-hour rental:

Company A:

A = 4(2) + 10

A = $18

Company B:

B = 6(2)

B = $12

For a 6-hour rental:

Company A:

A = 4(6) + 10

A = $34

Company B:

B = 6(6)

B = $36

The better price depends on how long the bicycle is rented.

Algebra allows us to investigate when the costs become equal.

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Finding the Break-Even Point

Set the two costs equal:

4h + 10 = 6h

Subtract 4h from both sides:

10 = 2h

Divide by 2:

h = 5

Therefore, both companies charge the same amount for:

5 hours

Check:

Company A:

4(5) + 10 = $30

Company B:

6(5) = $30

This value is sometimes called a break-even point.

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Algebra and Phone Plans

Phone and data plans often combine fixed and variable costs.

For example:

Plan A:

$20 monthly fee + $3 per GB

If:

g = additional gigabytes used

then:

Cost = 20 + 3g

Another plan might charge:

Cost = 35 + g

Algebra can help compare the plans at different levels of usage.

This is an example of algebra helping consumers make decisions.

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Algebra and Recipes

Recipes use mathematical relationships between quantities.

Suppose a recipe uses:

200 g of flour for one batch

Let:

b = number of batches

Then:

Flour needed = 200b

For 3 batches:

Flour = 200(3)

Flour = 600 g

For 2.5 batches:

Flour = 200(2.5)

Flour = 500 g

Algebra allows recipes to be scaled easily.

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Algebra and Geometry

Many geometric relationships are algebraic.

The area of a rectangle is:

A = lw

where:

• A = area
• l = length
• w = width

The circumference of a circle is:

C = 2πr

The area of a circle is:

A = πr²

These formulas are algebraic models that describe relationships between measurements.

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Worked Example — Garden

A rectangular garden has:

Width = x metres

and:

Length = x + 4 metres

Its area can be represented by:

A = x(x + 4)

If:

x = 6

then:

A = 6(6 + 4)

A = 6 × 10

A = 60 m²

The algebraic expression describes gardens of many possible sizes.

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Algebra and Temperature

Temperature conversions are another real-world application of algebra.

Celsius and Fahrenheit are related by:

F = 9C/5 + 32

where:

• C = temperature in degrees Celsius
• F = temperature in degrees Fahrenheit

For example, if:

C = 20

then:

F = 9(20)/5 + 32

F = 36 + 32

F = 68°F

A single algebraic equation allows any Celsius temperature to be converted to Fahrenheit.

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Algebra in Science

Science relies heavily on algebra.

Many scientific relationships are expressed using equations.

For example:

Density = Mass / Volume

or:

ρ = m/V

Newton's Second Law:

F = ma

Wave speed:

v = λf

Electrical relationships:

V = IR

These are all algebraic models.

Scientists use algebra to calculate unknown quantities and predict how systems will behave.

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Algebra and Predictions

One of the greatest advantages of algebra is that it allows us to make predictions.

Suppose a tank contains:

500 L of water

and loses:

20 L every hour

Let:

t = number of hours

The amount remaining is:

V = 500 − 20t

After 5 hours:

V = 500 − 20(5)

V = 400 L

After 10 hours:

V = 300 L

The equation allows us to predict the amount of water remaining at different times.

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Predicting When Something Will Happen

We can also use algebra to determine when a particular event will occur.

Using:

V = 500 − 20t

When will the tank contain only:

100 L?

Set:

V = 100

Then:

100 = 500 − 20t

Subtract 500:

−400 = −20t

Divide by −20:

t = 20

Therefore:

The tank will contain 100 L after 20 hours.

This is more useful than simply calculating one future value because it helps us plan ahead.

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Algebra and Graphs

Real-world algebraic relationships can often be represented using graphs.

Consider:

C = 5h + 10

where C is cost and h is hours.

The graph can show how the cost changes as time increases.

The starting value represents the fixed charge.

The steepness of the graph represents how quickly the cost increases.

Expressions, equations, tables, and graphs can therefore represent the same real-world relationship in different ways.

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Making Decisions with Algebra

Suppose you need to choose between two payment plans.

You could simply guess which one seems cheaper.

A better approach is to:

• Define variables.
• Create expressions for each plan.
• Calculate several examples.
• Find where the costs are equal.
• Compare the results.
• Consider which range applies to your situation.

This turns a decision into a mathematical problem.

Algebra does not necessarily make the decision for us. It provides evidence that helps us make an informed decision.

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Building an Algebraic Model

A useful process for modeling a real situation is:

Identify the quantities involved.

Ask which quantities are known and which can change.

Choose variables.

For example:

h = hours

Identify relationships.

Ask how one quantity depends on another.

Write an expression or equation.

For example:

C = 8h + 12

Use the model.

Substitute values, solve equations, or compare results.

Interpret the answer.

Connect the mathematical answer back to the original situation.

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Worked Example — Parking

A parking garage charges:

$3 entrance fee

and:

$2 per hour

Let:

h = number of hours parked

The cost is:

C = 2h + 3

For 4 hours:

C = 2(4) + 3

C = 11

Therefore:

4 hours of parking costs $11.

If you have $17 available:

17 = 2h + 3

Subtract 3:

14 = 2h

Divide by 2:

h = 7

Therefore:

$17 pays for 7 hours of parking.

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Worked Example — Saving Money

Suppose you already have:

$120

and save:

$30 each week

Let:

w = number of weeks

Total savings:

S = 120 + 30w

After 8 weeks:

S = 120 + 30(8)

S = 360

Therefore:

You will have $360 after 8 weeks.

If your goal is $600:

600 = 120 + 30w

Subtract 120:

480 = 30w

Divide by 30:

w = 16

Therefore:

It will take 16 weeks to reach $600.

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Worked Example — Fuel

Suppose a vehicle uses approximately:

7 L of fuel per 100 km

For a simplified model, let:

d = distance travelled in kilometres

Fuel used can be represented by:

F = 7d/100

If the vehicle travels:

350 km

then:

F = 7(350)/100

F = 24.5 L

If fuel costs $2 per litre:

Cost = 2(24.5)

Cost = $49

This shows how several algebraic relationships can be combined to solve a practical problem.

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Limitations of Algebraic Models

Real-world models are often simplifications.

For example:

d = 80t

assumes a vehicle travels at a constant speed of 80 km/h.

In reality:

• Traffic may slow the vehicle.
• The vehicle may stop.
• Road conditions may change.
• Speed limits may change.

Similarly, a fuel-consumption model may not account for:

• Driving style
• Traffic
• Weather
• Vehicle load
• Road conditions

A mathematical model can still be useful, but we should understand its assumptions.

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Common Mistakes

Mistake: Choosing a variable without defining it

Do not simply write:

C = 5x + 10

Explain what x represents.

For example:

x = number of hours

Mistake: Confusing fixed and variable costs

In:

C = 4h + 12

the 12 is the fixed cost.

The 4h changes depending on h.

Mistake: Forgetting units

A result of:

20

may mean:

• $20
• 20 km
• 20 hours
• 20 litres

Always interpret the answer in context.

Mistake: Creating an expression but not interpreting it

The expression:

5n + 8

has little practical meaning unless we know what:

n, 5, and 8

represent.

Mistake: Assuming every mathematical model is exact

Many real-world models make simplifying assumptions.

Mistake: Stopping after the calculation

If you calculate:

h = 6

you should explain what this means.

For example:

The bicycle must be rented for 6 hours.

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Key Terms

Variable – a symbol representing a quantity that can change.

Expression – a mathematical phrase containing numbers, variables, and operations.

Equation – a mathematical statement showing that two expressions are equal.

Model – a mathematical representation of a real system or situation.

Fixed cost – a cost that does not change with usage.

Variable cost – a cost that changes depending on quantity or usage.

Substitution – replacing a variable with a known value.

Prediction – using a mathematical relationship to estimate a future or unknown value.

Break-even point – a value at which two costs, incomes, or other quantities are equal.

Assumption – a condition accepted when constructing a model.

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Check Your Understanding

1. Give three examples of everyday situations that could be represented using algebra.

2. A sandwich costs $6. Let n represent the number of sandwiches purchased. Write an expression for the total cost.

3. A delivery company charges $5 plus $3 per kilometre. Let d represent distance in kilometres. Write an expression for the total cost.

4. Use your expression from Question 3 to calculate the cost of an 8 km delivery.

5. A worker earns $20 per hour. Write an expression for their earnings after h hours.

6. Use your expression to calculate their earnings after 7.5 hours.

7. A person has $300 and spends $25 each week. Write an expression for the amount remaining after w weeks.

8. Use your expression to calculate how much remains after 8 weeks.

9. A car travels at a constant speed of 90 km/h. Write an expression for the distance travelled after t hours.

10. How far will the car travel in 3.5 hours?

11. A parking garage charges according to:

C = 4h + 6

Explain what the numbers 4 and 6 could represent.

12. Plan A costs:

A = 5h + 10

Plan B costs:

B = 7h

Find the value of h at which the two plans cost the same.

13. Explain one way algebra can help someone make a financial decision.

14. Explain why a real-world algebraic model may not always predict the future perfectly.

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Key Takeaways

• Algebra is used extensively in shopping, travel, budgeting, wages, science, measurement, technology, and planning.
• Variables represent quantities that can change.
• Algebraic expressions can describe relationships between real-world quantities.
• Many practical models contain both fixed and variable quantities.
• Substitution allows us to calculate particular values from a general model.
• Equations allow us to work backward and determine unknown quantities.
• Algebra can be used to compare different choices, such as prices or payment plans.
• A break-even point occurs when two alternatives have equal values.
• Algebra allows us to make predictions about future values.
• Expressions, equations, tables, and graphs can represent the same relationship in different ways.
• Mathematical models often contain assumptions and may simplify complicated real-world situations.
• A calculated answer should always be interpreted in the context of the original problem.
• Algebra provides mathematical evidence that can help people make better-informed decisions.

Did You Know?

Many apps use algebra continuously without showing it to you. Navigation apps estimate travel time using relationships among distance, speed, and time, shopping websites calculate discounts and delivery costs, and financial apps use mathematical models to estimate savings and expenses. In many cases, the algebra is happening automatically in the background.