1. Introduction to Three-Variable Systems

Introduction

Many mathematical and real-world problems involve more than two unknown quantities.

For example, a school may sell three types of tickets, a business may manufacture three different products, or a scientist may combine three substances to produce a mixture.

In these situations, two variables may not be enough to describe all the unknown information.

A system of three-variable equations allows us to represent relationships involving three unknown quantities and determine values that satisfy all the equations simultaneously.

Three-variable systems extend the ideas already used in two-variable systems. We still define variables, construct equations, and look for values that make every equation true. However, the mathematics becomes more complex because we are working with three unknowns instead of two.

Three-variable systems are an important foundation for algebra, geometry, engineering, economics, chemistry, and more advanced mathematics.

What Is a Three-Variable System?

A three-variable system is a collection of equations involving three unknown quantities.

The variables are commonly represented by:

x, y, and z

For example:

x + y + z = 12

2x + y − z = 7

x − y + 2z = 7

These three equations form a system.

The goal is to find values of x, y, and z that satisfy all three equations at the same time.

A possible solution is written as an ordered triple:

(x, y, z)

For this system, the solution is:

(4, 5, 3)

We can verify this by substituting the values into each equation.

First equation:

4 + 5 + 3 = 12

Second equation:

2(4) + 5 − 3 = 10

This does not equal 7.

Therefore, (4, 5, 3) is not a solution to the system.

This illustrates an important principle: a solution must satisfy every equation, not just one or two of them.

We will return to the correct solution later.

Recognising Equations with Three Unknowns

An equation with three variables contains three different unknown quantities.

Examples include:

x + y + z = 20

2x − 3y + 4z = 15

5x + 2y − z = 8

3x + 6y + 2z = 30

Each equation contains x, y, and z.

However, an equation does not need to display all three variables explicitly to belong to a three-variable system.

For example:

x + y = 10

y + z = 12

x + z = 14

Together, these equations form a three-variable system because the system involves x, y, and z.

We can rewrite the first equation as:

x + y + 0z = 10

The coefficient of z is zero.

Identifying the variables

Consider:

4a + 2b − 3c = 18

The variables are:

• a

• b

• c

The coefficients are:

• 4 for a

• 2 for b

• −3 for c

The constant on the right side is 18.

Variable names do not have to be x, y, and z. They can represent quantities such as price, distance, time, or mass.

The General Form of a Linear Equation in Three Variables

A linear equation in three variables can be written as:

ax + by + cz = d

where:

• x, y, and z are variables.

• a, b, and c are coefficients.

• d is a constant.

• At least one of a, b, and c is nonzero.

For example:

2x + 3y − 4z = 12

Here:

a = 2

b = 3

c = −4

d = 12

The equation is linear because each variable appears only to the first power, and the variables are not multiplied together.

Linear and nonlinear equations

Three-variable systems can contain nonlinear equations, but this topic focuses on linear systems.

What Does a Solution Represent?

A solution to a three-variable system is an ordered triple of values that makes every equation true.

For example, consider:

x + y + z = 9

2x + y + z = 11

x + 2y + z = 12

We can test:

(x, y, z) = (2, 3, 4)

First equation:

2 + 3 + 4 = 9

Correct.

Second equation:

2(2) + 3 + 4 = 11

Correct.

Third equation:

2 + 2(3) + 4 = 12

Correct.

All three equations are satisfied.

Therefore:

(2, 3, 4) is a solution to the system.

The order matters

The ordered triple:

(2, 3, 4)

means:

x = 2

y = 3

z = 4

The triple (4, 3, 2) represents different values and is not necessarily a solution.

Always write the values in the order specified by the variables.

Why One Equation Is Usually Not Enough

Consider:

x + y + z = 12

Many different triples satisfy this equation.

Examples include:

(2, 4, 6)

(3, 5, 4)

(1, 1, 10)

(6, 3, 3)

Each triple adds to 12.

One equation generally does not provide enough information to determine three unknowns uniquely.

Adding a second equation

Suppose we also know:

x − y = 2

We now have:

x + y + z = 12

x − y = 2

There are still infinitely many real-number solutions.

For example:

(5, 3, 4)

(4, 2, 6)

(3, 1, 8)

Each triple satisfies both equations.

Adding a third independent equation

Suppose we also know:

y + z = 7

Our system becomes:

x + y + z = 12

x − y = 2

y + z = 7

The third equation provides additional information.

From:

x + y + z = 12

and:

y + z = 7

we obtain:

x = 5

Then:

5 − y = 2

y = 3

Finally:

3 + z = 7

z = 4

Therefore:

(x, y, z) = (5, 3, 4)

The system now has a unique solution.

Important qualification

Three equations do not automatically guarantee a unique solution.

Some equations may repeat information or contradict one another.

A three-variable system may have:

• Exactly one solution.

• No solution.

• Infinitely many solutions.

The number and independence of the equations determine the possible outcomes.

Comparing Two-Variable and Three-Variable Systems

Two-variable and three-variable systems follow the same general mathematical principles, but they differ in several important ways.

Two-variable example

x + y = 10

x − y = 2

Add the equations:

2x = 12

x = 6

Then:

y = 4

Solution:

(6, 4)

Three-variable example

x + y + z = 12

x − y = 2

y + z = 7

As shown earlier:

x = 5

y = 3

z = 4

Solution:

(5, 3, 4)

The central idea remains the same: find values that satisfy all equations simultaneously.

Graphical Meaning of Three-Variable Equations

A linear equation in two variables generally represents a straight line on a two-dimensional coordinate plane.

A linear equation in three variables represents a plane in three-dimensional space.

For example:

x + y + z = 6

represents a plane containing all points whose coordinates add to 6.

Some points on this plane include:

(6, 0, 0)

(0, 6, 0)

(0, 0, 6)

(2, 2, 2)

What is a plane?

A plane is a flat, two-dimensional surface extending infinitely in three-dimensional space.

A sheet of paper provides a useful physical analogy, although a mathematical plane has no thickness and extends without boundaries.

What happens when three planes intersect?

A system of three linear equations corresponds to three planes.

The solution consists of points that lie on all three planes simultaneously.

Depending on the arrangement, the system may have one solution, no solution, or infinitely many solutions.

A Unique Solution: Three Planes Meeting at One Point

Consider:

x = 2

y = 3

z = 4

Each equation represents a plane.

The plane x = 2 contains all points whose x-coordinate is 2.

The plane y = 3 contains all points whose y-coordinate is 3.

The plane z = 4 contains all points whose z-coordinate is 4.

All three planes share exactly one point:

(2, 3, 4)

This is a system with a unique solution.

Algebraic example

x + y + z = 9

2x + y + z = 11

x + 2y + z = 12

The unique solution is:

(2, 3, 4)

Graphically, the three planes share one common point.

No Solution: Planes with No Common Point

Consider:

x + y + z = 6

x + y + z = 10

The left sides are identical, but the right sides are different.

The same three values cannot add to both 6 and 10.

These equations represent parallel, distinct planes.

Adding a third equation cannot create a common point.

Therefore, the system has no solution.

Important idea

A system has no solution when there is no point that satisfies all its equations.

Not every inconsistent three-variable system consists of parallel planes. Three planes may also intersect pairwise without sharing a single common point.

Infinitely Many Solutions: Planes Sharing a Line

Consider:

x + y + z = 6

2x + 2y + 2z = 12

The second equation is twice the first.

Therefore, both equations represent the same plane.

Now add:

x − y = 0

This equation represents another plane.

The first and third planes intersect along a line.

Every point on that line satisfies all three equations.

For example:

x = y

and:

2x + z = 6

Let x = t.

Then:

y = t

z = 6 − 2t

The solutions have the form:

(t, t, 6 − 2t)

where t can be any real number.

Examples include:

(0, 0, 6)

(1, 1, 4)

(2, 2, 2)

(3, 3, 0)

There are infinitely many solutions.

Organising Information in Three-Variable Systems

One of the most important skills is translating written information into mathematical equations.

When three unknown quantities are involved, careful organisation becomes especially important.

Example: School Supplies

A school shop sells:

• Notebooks.

• Pens.

• Pencils.

A customer purchases:

2 notebooks, 3 pens, and 1 pencil for $15.

Another purchases:

1 notebook, 2 pens, and 4 pencils for $13.

A third purchases:

3 notebooks, 1 pen, and 2 pencils for $18.

We want to determine the price of each item.

Step 1: Define variables

n = price of one notebook in dollars

p = price of one pen in dollars

c = price of one pencil in dollars

Step 2: Organise the information

Step 3: Construct equations

First purchase:

2n + 3p + c = 15

Second purchase:

n + 2p + 4c = 13

Third purchase:

3n + p + 2c = 18

These equations form a three-variable system.

Step 4: Interpret the meaning

A solution (n, p, c) would give the individual prices of notebooks, pens, and pencils.

The system must satisfy all three purchases simultaneously.

We will solve this system in a later section.

Using Tables to Organise Information

Tables are particularly useful when word problems contain several categories.

For example, suppose a cinema sells adult, student, and child tickets.

Let:

a = number of adult tickets

s = number of student tickets

c = number of child tickets

If 200 tickets are sold:

a + s + c = 200

Suppose the prices are:

Adult: $15

Student: $10

Child: $6

If the total revenue is $2,200:

15a + 10s + 6c = 2200

Suppose 20 more adult tickets than child tickets were sold:

a − c = 20

We now have three equations:

a + s + c = 200

15a + 10s + 6c = 2200

a − c = 20

These equations represent three separate pieces of information.

The first describes the total number of tickets.

The second describes the total revenue.

The third describes the relationship between two ticket categories.

A useful habit is to identify what each equation represents before attempting to solve the system.

Worked Example: Finding Three Unknown Numbers

Three numbers have a sum of 24.

The first number is twice the second.

The third number is 3 greater than the second.

Find the three numbers.

Step 1: Define variables

x = first number

y = second number

z = third number

Step 2: Translate the statements

The sum is 24:

x + y + z = 24

The first is twice the second:

x = 2y

The third is 3 greater than the second:

z = y + 3

Step 3: Substitute

Substitute x = 2y and z = y + 3 into the first equation:

2y + y + (y + 3) = 24

4y + 3 = 24

4y = 21

y = 5.25

Step 4: Find the other values

x = 2(5.25) = 10.5

z = 5.25 + 3 = 8.25

Step 5: Verify

Sum:

10.5 + 5.25 + 8.25 = 24

First is twice the second:

10.5 = 2(5.25)

Third is 3 greater than the second:

8.25 = 5.25 + 3

Answer: The three numbers are 10.5, 5.25, and 8.25.

This example demonstrates that three-variable systems can be solved using substitution when two variables are easily expressed in terms of the third.

Worked Example: Three Types of Tickets

A school sells adult, student, and child tickets for a performance.

The prices are:

• Adult: $15

• Student: $10

• Child: $6

A total of 200 tickets are sold.

The total revenue is $2,200.

There are 20 more adult tickets than child tickets.

Find the number of each ticket type.

Step 1: Define variables

a = adult tickets

s = student tickets

c = child tickets

Step 2: Construct the system

Total tickets:

a + s + c = 200

Total revenue:

15a + 10s + 6c = 2200

Relationship:

a − c = 20

Step 3: Express a in terms of c

a = c + 20

Substitute into the total ticket equation:

(c + 20) + s + c = 200

s = 180 − 2c

Step 4: Substitute into the revenue equation

15(c + 20) + 10(180 − 2c) + 6c = 2200

Expand:

15c + 300 + 1800 − 20c + 6c = 2200

Combine:

c + 2100 = 2200

c = 100

Step 5: Find a and s

a = c + 20

a = 120

s = 180 − 2(100)

s = −20

Step 6: Interpret

The mathematical solution is:

a = 120

s = −20

c = 100

However, a negative number of student tickets is impossible.

Therefore, the conditions do not describe a valid real-world ticket sale.

Answer: There is no physically meaningful solution using nonnegative whole-number ticket quantities.

This is an important reminder that an algebraically consistent system may still produce values that violate contextual restrictions.

Worked Example: Solving a Three-Variable System by Elimination

Consider:

x + y + z = 12

2x − y + z = 7

x + 2y − z = 6

We want to determine x, y, and z.

Step 1: Label the equations

Equation A:

x + y + z = 12

Equation B:

2x − y + z = 7

Equation C:

x + 2y − z = 6

Step 2: Eliminate z from A and B

Subtract A from B:

(2x − y + z) − (x + y + z) = 7 − 12

x − 2y = −5

Call this Equation D.

Step 3: Eliminate z from A and C

Add A and C:

(x + y + z) + (x + 2y − z) = 12 + 6

2x + 3y = 18

Call this Equation E.

Step 4: Solve the two-variable system

We now have:

x − 2y = −5

2x + 3y = 18

Multiply the first equation by 2:

2x − 4y = −10

Subtract:

(2x + 3y) − (2x − 4y) = 18 − (−10)

7y = 28

y = 4

Step 5: Find x

x − 2(4) = −5

x − 8 = −5

x = 3

Step 6: Find z

Substitute into Equation A:

3 + 4 + z = 12

z = 5

Step 7: Verify

Equation A:

3 + 4 + 5 = 12

Equation B:

2(3) − 4 + 5 = 7

Equation C:

3 + 2(4) − 5 = 6

All three equations are satisfied.

Answer: (x, y, z) = (3, 4, 5).

Why elimination works

The original system has three unknowns.

By eliminating z from two pairs of equations, we create a smaller system containing only x and y.

After solving that system, we substitute the values back to find z.

This is the main strategy used for solving many three-variable linear systems.

Revisiting the Opening Example

At the beginning of these notes, we considered:

x + y + z = 12

2x + y − z = 7

x − y + 2z = 7

Let's determine the correct solution.

Step 1: Eliminate z

Add the first and second equations:

3x + 2y = 19

Multiply the first equation by 2:

2x + 2y + 2z = 24

Subtract the third equation:

x + 3y = 17

Step 2: Solve the resulting system

3x + 2y = 19

x + 3y = 17

From the second:

x = 17 − 3y

Substitute:

3(17 − 3y) + 2y = 19

51 − 9y + 2y = 19

−7y = −32

y = 32/7

Then:

x = 17 − 3(32/7)

x = 23/7

Step 3: Find z

x + y + z = 12

23/7 + 32/7 + z = 12

55/7 + z = 84/7

z = 29/7

Step 4: Verify

First equation:

23/7 + 32/7 + 29/7 = 84/7 = 12

Second equation:

2(23/7) + 32/7 − 29/7 = 49/7 = 7

Third equation:

23/7 − 32/7 + 2(29/7) = 49/7 = 7

Correct solution: (23/7, 32/7, 29/7).

Applications of Three-Variable Systems

Business and Economics

Businesses may need to determine the prices or quantities of three products using sales information.

Three-variable systems can also model production costs and financial relationships.

Chemistry

Scientists may use systems to determine the quantities of several substances in a mixture.

They can also use linear equations when balancing chemical reactions or analysing experimental data.

Engineering

Engineers use systems of equations to model forces, electrical circuits, structures, and mechanical systems.

Geometry

Three-variable equations represent planes in three-dimensional space.

Their intersections help describe geometric relationships.

Computer Science

Systems of equations are important in computer graphics, simulations, optimisation, and numerical modelling.

Transport and Logistics

Three-variable systems can describe quantities such as vehicle numbers, delivery costs, and transport capacities.

Common Mistakes

Assuming three variables always require exactly three equations

Three independent equations are typically sufficient for a unique solution, but three equations may be dependent or inconsistent. Some systems may require additional information.

Confusing an equation with a system

An equation is one mathematical relationship. A system contains multiple equations considered together.

Forgetting the order of the solution

The ordered triple (x, y, z) must match the variable definitions.

Checking only one equation

A valid solution must satisfy every equation.

Assuming all three-variable systems have one solution

A system may have one solution, no solution, or infinitely many solutions.

Incorrectly translating word problems

For example, "twice as many adults as children" means a = 2c, not a = c + 2.

Losing negative signs during elimination

When subtracting an equation, every term must be subtracted correctly.

Confusing three-dimensional coordinates with two-dimensional coordinates

A point in three-dimensional space has three coordinates, not two.

Accepting unrealistic contextual values

Negative numbers of tickets or fractional numbers of people may not make sense.

Failing to organise information

Tables, variable definitions, and labelled equations help prevent errors.

Key Terms

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

• Three-variable system: A system involving three unknown quantities.

• Variable: A symbol representing an unknown quantity.

• Coefficient: A number multiplying a variable.

• Constant: A fixed numerical value.

• Linear equation: An equation in which variables occur only to the first power and are not multiplied together.

• Solution: Values that satisfy all equations in a system.

• Ordered triple: Three values written in a specified order, such as (x, y, z).

• Independent equations: Equations that provide distinct mathematical information.

• Dependent equations: Equations containing redundant information.

• Consistent system: A system with at least one solution.

• Inconsistent system: A system with no solution.

• Unique solution: Exactly one solution.

• Infinitely many solutions: A system with an unlimited number of solutions.

• Plane: A flat, two-dimensional surface in three-dimensional space.

• Intersection: A point or set of points shared by geometric objects.

• Substitution: Replacing a variable with an equivalent expression.

• Elimination: Combining equations to remove a variable.

• Verification: Checking that proposed values satisfy the original equations.

• Constraint: A restriction on the possible values of variables.

Key Takeaways

• Three-variable systems involve three unknown quantities, commonly x, y, and z.

• A linear equation in three variables has the general form ax + by + cz = d.

• A solution is an ordered triple that satisfies every equation.

• One or two equations generally do not determine three unknowns uniquely.

• Three independent, consistent linear equations can determine a unique solution.

• A system may have one solution, no solution, or infinitely many solutions.

• Linear equations in three variables represent planes in three-dimensional space.

• Tables and clearly defined variables help organise information from word problems.

• Elimination can reduce a three-variable system to a two-variable system.

• Every solution should be verified by substitution into the original equations.

• Contextual restrictions must be considered when interpreting real-world answers.

Check Your Understanding

Recall and Understanding

1. What is a three-variable system?

2. Write the general form of a linear equation in three variables.

3. What does the ordered triple (2, 5, 7) represent?

4. Explain why one linear equation in three variables generally does not determine a unique solution.

5. Describe the graphical difference between two-variable and three-variable linear systems.

Identifying and Organising Systems

6. Identify the variables and coefficients in:

3x − 2y + 5z = 18

7. Determine whether the following equations are linear:

a. x + 2y − z = 9

b. x² + y + z = 10

c. 3x + 4y = 12

d. xy + z = 7

8. A shop sells three types of stationery. Two notebooks, one pen, and three pencils cost $14. One notebook, three pens, and two pencils cost $13. Three notebooks, two pens, and one pencil cost $19. Define variables and construct the system.

9. Explain the difference between a consistent and an inconsistent system.

10. Can three linear equations in three variables have infinitely many solutions? Explain.

Application

11. Determine whether (2, 3, 4) satisfies:

x + y + z = 9

2x + y − z = 3

x − y + 2z = 7

12. Solve:

x + y + z = 15

x − y = 3

y + z = 9

13. Solve:

x + y + z = 10

2x + y + z = 13

x + 2y + z = 14

14. Determine whether the system has a solution:

x + y + z = 8

2x + 2y + 2z = 20

x − y = 3

15. A school sells adult, student, and child tickets. A total of 100 tickets are sold. Adult tickets cost $12, student tickets cost $8, and child tickets cost $5. The total revenue is $850. There are 10 more adult tickets than child tickets. Construct the system and determine the number of each ticket type.

Answers to Check Your Understanding

1. A three-variable system is a collection of equations involving three unknown quantities that must be satisfied simultaneously.

2.

ax + by + cz = d

At least one of a, b, and c must be nonzero.

3.

x = 2

y = 5

z = 7

The order of the values matters.

4. One equation usually allows infinitely many combinations of three values. Additional independent relationships are needed to determine a unique solution.

5. A linear equation in two variables represents a line in two-dimensional space. A linear equation in three variables represents a plane in three-dimensional space.

6.

Variables:

x, y, z

Coefficients:

3, −2, 5

Constant:

18

7.

a. Linear

b. Nonlinear, because x is squared.

c. Linear; the coefficient of z is zero.

d. Nonlinear, because x and y are multiplied.

8.

Let:

n = price of one notebook

p = price of one pen

c = price of one pencil

The system is:

2n + p + 3c = 14

n + 3p + 2c = 13

3n + 2p + c = 19

9. A consistent system has at least one solution. An inconsistent system has no solution.

10. Yes. If the equations are dependent or share a common line or plane, infinitely many solutions may exist.

11.

Test (2, 3, 4).

First equation:

2 + 3 + 4 = 9

Correct.

Second equation:

2(2) + 3 − 4 = 3

Correct.

Third equation:

2 − 3 + 2(4) = 7

Correct.

Answer: Yes, (2, 3, 4) satisfies the system.

12.

x + y + z = 15

x − y = 3

y + z = 9

Subtract the third equation from the first:

x = 6

Then:

6 − y = 3

y = 3

Finally:

3 + z = 9

z = 6

Answer: (6, 3, 6).

Verification:

6 + 3 + 6 = 15

6 − 3 = 3

3 + 6 = 9

13.

x + y + z = 10

2x + y + z = 13

x + 2y + z = 14

Subtract the first equation from the second:

x = 3

Subtract the first equation from the third:

y = 4

Substitute:

3 + 4 + z = 10

z = 3

Answer: (3, 4, 3).

14.

First equation:

x + y + z = 8

Multiply by 2:

2x + 2y + 2z = 16

However, the second equation states:

2x + 2y + 2z = 20

This is a contradiction.

Answer: The system has no solution.

15.

Let:

a = adult tickets

s = student tickets

c = child tickets

The equations are:

a + s + c = 100

12a + 8s + 5c = 850

a − c = 10

From the third equation:

a = c + 10

Substitute into the first:

c + 10 + s + c = 100

s = 90 − 2c

Substitute into the revenue equation:

12(c + 10) + 8(90 − 2c) + 5c = 850

12c + 120 + 720 − 16c + 5c = 850

c + 840 = 850

c = 10

Then:

a = 20

s = 90 − 20 = 70

Verification:

20 + 70 + 10 = 100

12(20) + 8(70) + 5(10)

= 240 + 560 + 50

= $850

Adult tickets exceed child tickets by:

20 − 10 = 10

Answer: 20 adult tickets, 70 student tickets, and 10 child tickets.

Final Summary

Three-variable systems extend the principles of simultaneous equations to situations involving three unknown quantities.

These systems can be represented algebraically using equations such as ax + by + cz = d, and graphically using planes in three-dimensional space.

A solution is an ordered triple that satisfies every equation. Depending on the relationships between the equations, a system may have one solution, no solution, or infinitely many solutions.

The central idea is that each equation provides information about the unknown quantities, and a valid solution must satisfy all the information simultaneously.

2. Solving Three-Variable Systems

Introduction

Many mathematical and real-world problems involve three unknown quantities connected by several relationships. These problems can be represented using systems of three linear equations.

For example, a school might need to determine the prices of three different products, a business might calculate the quantities of three items sold, or an engineer might determine three unknown forces acting on a structure.

To solve these problems, we need to find values for all three variables that satisfy every equation in the system.

One of the most effective methods is elimination.

Elimination allows us to simplify a complicated system by removing one variable at a time. We begin with three equations containing three unknowns, reduce them to two equations containing two unknowns, and then solve the resulting system.

Once two variables have been found, we substitute their values into one of the original equations to determine the third variable.

The key to success is working systematically, organising each step clearly, and verifying the final solution.

Reviewing Three-Variable Systems

A linear equation in three variables has the general form:

ax + by + cz = d

where:

• x, y, and z are variables.

• a, b, and c are coefficients.

• d is a constant.

• At least one of a, b, and c is nonzero.

A system of three linear equations may be written as:

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

The goal is to determine an ordered triple:

(x, y, z)

that satisfies all three equations simultaneously.

Example

Consider:

x + y + z = 12

2x − y + z = 7

x + 2y − z = 6

The solution is:

(x, y, z) = (3, 4, 5)

Verification:

First equation:

3 + 4 + 5 = 12

Second equation:

2(3) − 4 + 5 = 7

Third equation:

3 + 2(4) − 5 = 6

All three equations are satisfied.

Why Elimination Works

Elimination uses the addition or subtraction of equations to remove a variable.

The principle is the same as in two-variable systems.

For example:

x + y = 10

x − y = 4

Adding the equations eliminates y:

2x = 14

x = 7

Then:

y = 3

In a three-variable system, we use the same idea, but we must eliminate a variable from two different pairs of equations.

This produces two equations containing only two unknowns.

We then solve that smaller system.

The elimination process

The process is sometimes called successive elimination because variables are removed in stages.

The General Elimination Method

To solve a system of three linear equations:

1. Label the original equations A, B, and C.

2. Choose one variable to eliminate first.

3. Eliminate that variable using two of the equations.

4. Eliminate the same variable using another pair of equations.

5. Solve the resulting two-variable system.

6. Substitute the known values into an original equation to find the third variable.

7. Check the ordered triple in all three original equations.

Important rule

When eliminating the first variable, use the same variable in both pairs.

For example, if you eliminate z from Equations A and B, you should also eliminate z from another pair, such as A and C.

Otherwise, the two resulting equations may contain different sets of variables and will not form the intended two-variable system.

Worked Example: Elimination Without Multiplication

Solve:

x + y + z = 12

2x − y + z = 7

x + 2y − z = 6

Step 1: Label the equations

A: x + y + z = 12

B: 2x − y + z = 7

C: x + 2y − z = 6

Step 2: Eliminate z from A and B

Both equations contain +z.

Subtract A from B:

(2x − y + z) − (x + y + z) = 7 − 12

Simplify:

x − 2y = −5

Call this Equation D.

Step 3: Eliminate z from A and C

Equation A contains +z.

Equation C contains −z.

Add the equations:

(x + y + z) + (x + 2y − z) = 12 + 6

Simplify:

2x + 3y = 18

Call this Equation E.

Step 4: Solve the two-variable system

We now have:

D: x − 2y = −5

E: 2x + 3y = 18

Multiply D by 2:

2x − 4y = −10

Subtract this from E:

(2x + 3y) − (2x − 4y) = 18 − (−10)

7y = 28

y = 4

Step 5: Find x

Substitute y = 4 into D:

x − 2(4) = −5

x − 8 = −5

x = 3

Step 6: Find z

Substitute x = 3 and y = 4 into A:

3 + 4 + z = 12

z = 5

Step 7: Verify

A:

3 + 4 + 5 = 12

B:

2(3) − 4 + 5 = 7

C:

3 + 2(4) − 5 = 6

All three equations are satisfied.

Final answer: (3, 4, 5).

Organising Elimination Work

Three-variable systems often involve many algebraic steps.

A clear layout reduces errors.

For example:

Labelling equations is particularly helpful when coefficients are large or fractions are involved.

Worked Example: Multiplying Equations Before Elimination

Sometimes the coefficients of the variable we want to eliminate are not equal or opposite.

In these situations, we must multiply one or both equations by suitable constants.

Solve:

2x + 3y + z = 13

3x − 2y + 2z = 9

x + y − z = 1

Step 1: Label the equations

A: 2x + 3y + z = 13

B: 3x − 2y + 2z = 9

C: x + y − z = 1

We will eliminate z first.

Step 2: Eliminate z from A and B

Equation A contains +z.

Equation B contains +2z.

Multiply A by 2:

4x + 6y + 2z = 26

Subtract B:

(4x + 6y + 2z) − (3x − 2y + 2z) = 26 − 9

Simplify:

x + 8y = 17

Call this Equation D.

Step 3: Eliminate z from A and C

Equation A contains +z.

Equation C contains −z.

Add A and C:

(2x + 3y + z) + (x + y − z) = 13 + 1

Simplify:

3x + 4y = 14

Call this Equation E.

Step 4: Solve D and E

D: x + 8y = 17

E: 3x + 4y = 14

Multiply D by 3:

3x + 24y = 51

Subtract E:

20y = 37

y = 37/20

Step 5: Find x

Substitute into D:

x + 8(37/20) = 17

x + 74/5 = 17

x = 11/5

Step 6: Find z

Substitute into C:

x + y − z = 1

11/5 + 37/20 − z = 1

44/20 + 37/20 − z = 20/20

81/20 − z = 20/20

z = 61/20

Step 7: Verify

A:

2(11/5) + 3(37/20) + 61/20

= 88/20 + 111/20 + 61/20

= 260/20 = 13

B:

3(11/5) − 2(37/20) + 2(61/20)

= 132/20 − 74/20 + 122/20

= 180/20 = 9

C:

11/5 + 37/20 − 61/20

= 44/20 + 37/20 − 61/20

= 20/20 = 1

Final answer: (11/5, 37/20, 61/20).

This example demonstrates why fractions should be kept exact during multi-step algebra.

Rounding too early can produce inaccurate results.

Choosing Which Variable to Eliminate

Any variable can be eliminated first, but some choices make the calculation easier.

Consider:

x + 2y + z = 11

2x − y + z = 7

3x + y − z = 8

The coefficients of z are:

+1, +1, −1

These are convenient because z can be eliminated by simple addition or subtraction.

In contrast, eliminating x would require working with coefficients 1, 2, and 3.

Useful strategies

Choose a variable that:

• Has equal coefficients in two equations.

• Has opposite coefficients in two equations.

• Has coefficients of 1 or −1.

• Can be eliminated using small multipliers.

The goal is to reduce unnecessary arithmetic.

Worked Example: Choosing the Efficient Variable

Solve:

x + 2y + z = 11

2x − y + z = 7

3x + y − z = 8

Step 1: Eliminate z from the first two equations

Subtract the first equation from the second:

(2x − y + z) − (x + 2y + z) = 7 − 11

x − 3y = −4

Call this D.

Step 2: Eliminate z from the first and third equations

Add the equations:

(x + 2y + z) + (3x + y − z) = 11 + 8

4x + 3y = 19

Call this E.

Step 3: Solve D and E

D: x − 3y = −4

E: 4x + 3y = 19

Add:

5x = 15

x = 3

Substitute into D:

3 − 3y = −4

−3y = −7

y = 7/3

Step 4: Find z

Substitute into the first equation:

3 + 2(7/3) + z = 11

3 + 14/3 + z = 11

9/3 + 14/3 + z = 33/3

z = 10/3

Step 5: Verify

First equation:

3 + 14/3 + 10/3 = 11

Second equation:

2(3) − 7/3 + 10/3 = 7

Third equation:

3(3) + 7/3 − 10/3 = 8

Final answer: (3, 7/3, 10/3).

Worked Example: Eliminating x Instead of z

Elimination does not always have to begin with z.

Solve:

x + y + z = 9

x + 2y + 3z = 17

x − y + 2z = 7

Step 1: Label equations

A: x + y + z = 9

B: x + 2y + 3z = 17

C: x − y + 2z = 7

All three equations have a coefficient of 1 for x.

Therefore, eliminating x is convenient.

Step 2: Eliminate x from A and B

Subtract A from B:

y + 2z = 8

Call this D.

Step 3: Eliminate x from A and C

Subtract A from C:

−2y + z = −2

Call this E.

Step 4: Solve D and E

D: y + 2z = 8

E: −2y + z = −2

Multiply D by 2:

2y + 4z = 16

Add E:

5z = 14

z = 14/5

Step 5: Find y

y + 2(14/5) = 8

y + 28/5 = 40/5

y = 12/5

Step 6: Find x

x + 12/5 + 14/5 = 9

x + 26/5 = 45/5

x = 19/5

Step 7: Verify

A:

19/5 + 12/5 + 14/5 = 45/5 = 9

B:

19/5 + 24/5 + 42/5 = 85/5 = 17

C:

19/5 − 12/5 + 28/5 = 35/5 = 7

Final answer: (19/5, 12/5, 14/5).

Elimination with Negative Coefficients

Negative coefficients require careful attention.

Consider:

2x − 3y + z = 4

−x + 2y − z = 1

3x + y + 2z = 14

We can eliminate z by adding the first two equations.

Step 1: Add the first two equations

(2x − 3y + z) + (−x + 2y − z) = 4 + 1

x − y = 5

Call this D.

Step 2: Eliminate z from the first and third equations

Multiply the first equation by 2:

4x − 6y + 2z = 8

Subtract this from the third equation:

(3x + y + 2z) − (4x − 6y + 2z) = 14 − 8

−x + 7y = 6

Call this E.

Step 3: Solve D and E

D: x − y = 5

E: −x + 7y = 6

Add:

6y = 11

y = 11/6

Then:

x = 5 + 11/6

x = 41/6

Step 4: Find z

Use the first equation:

2(41/6) − 3(11/6) + z = 4

82/6 − 33/6 + z = 24/6

49/6 + z = 24/6

z = −25/6

Step 5: Verify

First equation:

2(41/6) − 3(11/6) − 25/6

= 82/6 − 33/6 − 25/6

= 24/6 = 4

Second equation:

−41/6 + 2(11/6) − (−25/6)

= −41/6 + 22/6 + 25/6

= 6/6 = 1

Third equation:

3(41/6) + 11/6 + 2(−25/6)

= 123/6 + 11/6 − 50/6

= 84/6 = 14

Final answer: (41/6, 11/6, −25/6).

Notice that a negative value for z is perfectly acceptable in an abstract algebraic system.

Whether negative values are reasonable in a real-world problem depends on what the variables represent.

Organising Multi-Step Solutions

Three-variable elimination can become difficult when equations are written without a clear structure.

A useful method is to label each equation and keep the reduced system separate from the original system.

Recommended layout

Original equations:

A: 2x + y − z = 8

B: x − y + 2z = 3

C: 3x + 2y + z = 14

First elimination:

D: Equation obtained from A and B

E: Equation obtained from A and C

Reduced system:

D: Equation in two variables

E: Equation in two variables

Solve:

Find the first variable.

Find the second variable.

Back-substitute:

Find the third variable.

Verification:

Check A, B, and C.

This approach makes it easier to identify where a calculation error occurred.

Worked Example: Solving a System with Larger Coefficients

Solve:

2x + 3y + 4z = 25

3x − 2y + z = 10

x + 4y − 2z = 5

Step 1: Label equations

A: 2x + 3y + 4z = 25

B: 3x − 2y + z = 10

C: x + 4y − 2z = 5

We will eliminate z.

Step 2: Eliminate z from A and B

Multiply B by 4:

12x − 8y + 4z = 40

Subtract A:

10x − 11y = 15

Call this D.

Step 3: Eliminate z from A and C

Multiply C by 2:

2x + 8y − 4z = 10

Add A:

4x + 11y = 35

Call this E.

Step 4: Solve D and E

D: 10x − 11y = 15

E: 4x + 11y = 35

Add:

14x = 50

x = 25/7

Step 5: Find y

Substitute into E:

4(25/7) + 11y = 35

100/7 + 11y = 245/7

11y = 145/7

y = 145/77

Step 6: Find z

Use B:

3x − 2y + z = 10

3(25/7) − 2(145/77) + z = 10

825/77 − 290/77 + z = 770/77

535/77 + z = 770/77

z = 235/77

Step 7: Verify

A:

2(25/7) + 3(145/77) + 4(235/77)

= 550/77 + 435/77 + 940/77

= 1925/77 = 25

B:

3(25/7) − 2(145/77) + 235/77

= 825/77 − 290/77 + 235/77

= 770/77 = 10

C:

25/7 + 4(145/77) − 2(235/77)

= 275/77 + 580/77 − 470/77

= 385/77 = 5

Final answer: (25/7, 145/77, 235/77).

This example shows that elimination remains effective even when the solution contains fractions.

Using Elimination in Word Problems

Three-variable systems are useful when a situation contains three unknown quantities and enough independent information to determine them.

Worked Example: Prices of Three Products

A school shop sells notebooks, pens, and pencils.

Two notebooks, one pen, and one pencil cost $13.

One notebook, two pens, and three pencils cost $14.

Three notebooks, one pen, and two pencils cost $20.

Find the price of each item.

Step 1: Define variables

n = price of one notebook

p = price of one pen

c = price of one pencil

Step 2: Construct equations

A: 2n + p + c = 13

B: n + 2p + 3c = 14

C: 3n + p + 2c = 20

Step 3: Eliminate p from A and C

Subtract A from C:

n + c = 7

Call this D.

Step 4: Eliminate p from A and B

Multiply A by 2:

4n + 2p + 2c = 26

Subtract B:

3n − c = 12

Call this E.

Step 5: Solve D and E

D: n + c = 7

E: 3n − c = 12

Add:

4n = 19

n = 4.75

Then:

c = 7 − 4.75

c = 2.25

Step 6: Find p

Use A:

2(4.75) + p + 2.25 = 13

9.50 + p + 2.25 = 13

p = 1.25

Step 7: Verify

A:

2(4.75) + 1.25 + 2.25 = 13

B:

4.75 + 2(1.25) + 3(2.25)

= 4.75 + 2.50 + 6.75

= 14

C:

3(4.75) + 1.25 + 2(2.25)

= 14.25 + 1.25 + 4.50

= 20

Answer: Notebook = $4.75; pen = $1.25; pencil = $2.25.

Worked Example: Three Types of Tickets

A theatre sells adult, student, and child tickets.

Adult tickets cost $15.

Student tickets cost $10.

Child tickets cost $5.

A total of 100 tickets are sold.

The total revenue is $1,050.

There are twice as many student tickets as child tickets.

Find the number of each type of ticket.

Step 1: Define variables

a = adult tickets

s = student tickets

c = child tickets

Step 2: Construct equations

Total tickets:

a + s + c = 100

Total revenue:

15a + 10s + 5c = 1050

Relationship:

s = 2c

Step 3: Eliminate a

Multiply the total ticket equation by 15:

15a + 15s + 15c = 1500

Subtract the revenue equation:

5s + 10c = 450

Divide by 5:

s + 2c = 90

Call this D.

Step 4: Use the third equation

We also know:

s = 2c

Substitute into D:

2c + 2c = 90

4c = 90

c = 22.5

Then:

s = 45

And:

a = 100 − 45 − 22.5

a = 32.5

Step 5: Interpret

The algebraic solution is:

a = 32.5

s = 45

c = 22.5

However, tickets must be counted in whole numbers.

Therefore, the original conditions have no valid whole-number solution.

This is a useful reminder that solving a system algebraically does not guarantee that its solution is realistic.

Recognising Special Cases

Three-variable systems do not always have exactly one solution.

During elimination, we may encounter an equation that reveals something important about the system.

No solution

Consider:

x + y + z = 10

2x + 2y + 2z = 25

x − y + z = 4

Multiply the first equation by 2:

2x + 2y + 2z = 20

But the second equation states:

2x + 2y + 2z = 25

Subtracting gives:

0 = 5

This is impossible.

Therefore, the system is inconsistent.

Answer: No solution.

Infinitely many solutions

Consider:

x + y + z = 10

2x + 2y + 2z = 20

x − y = 2

The second equation is twice the first.

It provides no new independent information.

From:

x − y = 2

we obtain:

x = y + 2

Substitute into the first equation:

(y + 2) + y + z = 10

2y + z = 8

z = 8 − 2y

Let y = t.

Then:

x = t + 2

y = t

z = 8 − 2t

The solution set is:

(t + 2, t, 8 − 2t)

where t is any real number.

There are infinitely many solutions.

Unique solution

If elimination produces enough independent equations to determine one value for each variable, the system has a unique solution.

Verifying Solutions

Verification is essential because three-variable systems involve several calculations and opportunities for errors.

Suppose we obtain:

x = 2

y = 3

z = 4

for the system:

x + y + z = 9

2x + y − z = 3

x − y + 2z = 7

Check the first equation:

2 + 3 + 4 = 9

Correct.

Check the second equation:

2(2) + 3 − 4 = 3

Correct.

Check the third equation:

2 − 3 + 2(4) = 7

Correct.

Since all three equations are satisfied, the solution is verified.

Why checking one equation is insufficient

A proposed triple may satisfy one or two equations but fail the third.

For example:

(3, 2, 4)

satisfies:

x + y + z = 9

because:

3 + 2 + 4 = 9

However, it does not satisfy:

2x + y − z = 3

because:

2(3) + 2 − 4 = 4

Therefore, it is not a solution to the complete system.

Checking Answers in Context

When solving real-world problems, verification involves more than substitution.

We must also consider whether the values make sense.

For example:

• Prices should generally be nonnegative.

• Numbers of people should be whole numbers.

• Distances should use appropriate units.

• Masses and volumes should not be negative.

• Quantities should satisfy any limits stated in the problem.

A system may have a mathematically correct solution that is invalid in the original context.

Common Mistakes

Eliminating different variables from different pairs

When reducing a three-variable system, eliminate the same variable from two different pairs to obtain two equations in the same remaining variables.

Forgetting to multiply every term

If an equation is multiplied by 3, every coefficient and the constant must be multiplied by 3.

Incorrect subtraction of negative terms

For example:

5y − (−2y) = 7y

not 3y.

Losing track of equations

Label equations A, B, C, D, and E to keep the work organised.

Stopping after finding two variables

A complete solution must include all three variables.

Substituting into a reduced equation that no longer contains the third variable

To find the eliminated variable, return to an original equation containing it.

Rounding too early

Keep fractions exact until the final answer when possible.

Assuming every system has a unique solution

Some systems have no solution or infinitely many solutions.

Not verifying the result

Substitute the final ordered triple into all three original equations.

Ignoring contextual restrictions

A mathematically valid result may not make sense in a word problem.

Key Terms

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

• Three-variable system: A system involving three unknown quantities.

• Linear equation: An equation in which variables appear only to the first power and are not multiplied together.

• Elimination: A method of removing variables by combining equations.

• Successive elimination: Eliminating variables in stages to simplify a system.

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

• Back-substitution: Using previously calculated values to find remaining unknowns.

• Coefficient: A number multiplying a variable.

• Constant: A fixed numerical value.

• Ordered triple: Three values written in a specified order, such as (x, y, z).

• Unique solution: Exactly one set of values satisfies the system.

• Consistent system: A system with at least one solution.

• Inconsistent system: A system with no solution.

• Dependent equations: Equations that provide redundant information.

• Contradiction: An impossible statement such as 0 = 5.

• Identity: A statement true for all relevant values, such as 0 = 0.

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

• Constraint: A restriction on the possible values of variables.

Key Takeaways

• Elimination is an effective method for solving three-variable systems.

• The goal is to reduce three equations in three variables to two equations in two variables.

• The same variable should be eliminated from two different pairs of equations.

• Multiplication may be required to create matching or opposite coefficients.

• Labelling equations helps organise multi-step solutions.

• After solving the reduced system, back-substitution determines the third variable.

• Every solution should be written as an ordered triple.

• A system may have one solution, no solution, or infinitely many solutions.

• Contradictions indicate inconsistency.

• Redundant equations may lead to infinitely many solutions when the system remains consistent.

• Verification requires substituting the final values into all original equations.

• Contextual restrictions must be considered in real-world applications.

Check Your Understanding

Recall and Understanding

1. Explain why elimination is useful for solving systems of three linear equations.

2. Why should the same variable be eliminated from two different pairs of equations?

3. What is back-substitution, and why is it necessary?

4. What does the equation 0 = 8 indicate when it appears during elimination?

5. Explain why a solution must be checked in all three original equations.

Basic Elimination

6. Solve:

x + y + z = 12

x + y − z = 4

x − y + z = 6

7. Solve:

x + y + z = 15

2x + y + z = 19

x + 2y + z = 20

8. Solve:

2x + y + z = 13

x + y − z = 2

x − y + z = 4

9. Solve:

x + 2y + z = 14

2x − y + z = 8

x + y − z = 2

10. Solve:

2x + 3y − z = 7

x − y + z = 5

3x + 2y + z = 16

Application and Challenge

11. Solve:

2x + y + 3z = 17

x − 2y + z = 1

3x + y − z = 4

12. Determine whether the following system has a solution:

x + y + z = 9

2x + 2y + 2z = 20

x − y + z = 3

13. Determine whether the following system has one solution or infinitely many solutions:

x + y + z = 8

2x + 2y + 2z = 16

x − y = 2

14. A school shop sells notebooks, pens, and pencils.

• Two notebooks, one pen, and one pencil cost $14.

• One notebook, two pens, and three pencils cost $15.

• Three notebooks, two pens, and one pencil cost $23.

Find the price of each item.

15. A theatre sells adult, student, and child tickets for $15, $10, and $5 respectively. A total of 100 tickets are sold, generating $1,050. There are twice as many child tickets as adult tickets.

Construct and solve a three-variable system to determine how many of each ticket type were sold.

Answers to Check Your Understanding

1. Elimination simplifies a three-variable system by removing one variable from two pairs of equations. This produces a two-variable system that can be solved using familiar methods.

2. Eliminating the same variable produces two equations involving the same two remaining unknowns. These can then be solved simultaneously.

3. Back-substitution involves substituting known variable values into an earlier equation to find an unknown variable. It is necessary to recover variables that were eliminated.

4. The statement 0 = 8 is a contradiction. It indicates that the system is inconsistent and has no solution.

5. A proposed ordered triple must satisfy every equation simultaneously. Checking only one or two equations does not establish that it is a solution to the complete system.

6.

A: x + y + z = 12

B: x + y − z = 4

C: x − y + z = 6

Subtract B from A:

2z = 8

z = 4

Subtract C from A:

2y = 6

y = 3

Substitute into A:

x + 3 + 4 = 12

x = 5

Answer: (5, 3, 4).

Verification:

5 + 3 + 4 = 12

5 + 3 − 4 = 4

5 − 3 + 4 = 6

7.

A: x + y + z = 15

B: 2x + y + z = 19

C: x + 2y + z = 20

Subtract A from B:

x = 4

Subtract A from C:

y = 5

Substitute into A:

4 + 5 + z = 15

z = 6

Answer: (4, 5, 6).

Verification:

4 + 5 + 6 = 15

2(4) + 5 + 6 = 19

4 + 2(5) + 6 = 20

8.

A: 2x + y + z = 13

B: x + y − z = 2

C: x − y + z = 4

Add B and C:

2x = 6

x = 3

Add A and B:

3x + 2y = 15

Substitute x = 3:

9 + 2y = 15

y = 3

Use B:

3 + 3 − z = 2

z = 4

Answer: (3, 3, 4).

Verification:

2(3) + 3 + 4 = 13

3 + 3 − 4 = 2

3 − 3 + 4 = 4

9.

A: x + 2y + z = 14

B: 2x − y + z = 8

C: x + y − z = 2

Subtract B from A:

−x + 3y = 6

Add A and C:

2x + 3y = 16

Subtract the first reduced equation from the second:

3x = 10

x = 10/3

Substitute into:

−x + 3y = 6

−10/3 + 3y = 6

3y = 28/3

y = 28/9

Use C:

10/3 + 28/9 − z = 2

z = 40/9

Answer: (10/3, 28/9, 40/9).

Verification:

A: 10/3 + 2(28/9) + 40/9 = 14

B: 2(10/3) − 28/9 + 40/9 = 8

C: 10/3 + 28/9 − 40/9 = 2

10.

A: 2x + 3y − z = 7

B: x − y + z = 5

C: 3x + 2y + z = 16

Add A and B:

3x + 2y = 12

Subtract B from C:

2x + 3y = 11

Multiply the first reduced equation by 3:

9x + 6y = 36

Multiply the second by 2:

4x + 6y = 22

Subtract:

5x = 14

x = 14/5

Substitute into:

3x + 2y = 12

42/5 + 2y = 12

2y = 18/5

y = 9/5

Use B:

14/5 − 9/5 + z = 5

z = 4

Answer: (14/5, 9/5, 4).

Verification:

A: 28/5 + 27/5 − 4 = 7

B: 14/5 − 9/5 + 4 = 5

C: 42/5 + 18/5 + 4 = 16

11.

A: 2x + y + 3z = 17

B: x − 2y + z = 1

C: 3x + y − z = 4

Eliminate y.

Subtract A from C:

x − 4z = −13

Call this D.

Multiply A by 2:

4x + 2y + 6z = 34

Add B:

5x + 7z = 35

Call this E.

From D:

x = 4z − 13

Substitute into E:

5(4z − 13) + 7z = 35

20z − 65 + 7z = 35

27z = 100

z = 100/27

Then:

x = 4(100/27) − 13

x = 49/27

Use B:

49/27 − 2y + 100/27 = 1

149/27 − 2y = 27/27

2y = 122/27

y = 61/27

Answer: (49/27, 61/27, 100/27).

Verification:

A: (98 + 61 + 300)/27 = 459/27 = 17

B: (49 − 122 + 100)/27 = 27/27 = 1

C: (147 + 61 − 100)/27 = 108/27 = 4

12.

First equation:

x + y + z = 9

Multiply by 2:

2x + 2y + 2z = 18

But the second equation states:

2x + 2y + 2z = 20

Subtracting gives:

0 = 2

This is impossible.

Answer: No solution. The system is inconsistent.

13.

The system is:

x + y + z = 8

2x + 2y + 2z = 16

x − y = 2

The second equation is twice the first, so it provides no new information.

From:

x − y = 2

x = y + 2

Substitute into the first equation:

(y + 2) + y + z = 8

2y + z = 6

z = 6 − 2y

Let y = t.

Then:

x = t + 2

y = t

z = 6 − 2t

Answer: Infinitely many solutions, given by (t + 2, t, 6 − 2t), for any real number t.

14.

Let:

n = notebook price

p = pen price

c = pencil price

The equations are:

A: 2n + p + c = 14

B: n + 2p + 3c = 15

C: 3n + 2p + c = 23

Subtract A from C:

n + p = 9

Therefore:

p = 9 − n

Substitute into A:

2n + (9 − n) + c = 14

n + c = 5

Therefore:

c = 5 − n

Substitute both expressions into B:

n + 2(9 − n) + 3(5 − n) = 15

n + 18 − 2n + 15 − 3n = 15

33 − 4n = 15

4n = 18

n = 4.50

Then:

p = 9 − 4.50 = 4.50

c = 5 − 4.50 = 0.50

Answer: Notebook = $4.50; pen = $4.50; pencil = $0.50.

Verification:

2(4.50) + 4.50 + 0.50 = 14

4.50 + 2(4.50) + 3(0.50) = 15

3(4.50) + 2(4.50) + 0.50 = 23

15.

Let:

a = adult tickets

s = student tickets

c = child tickets

Construct the equations.

Total tickets:

a + s + c = 100

Total revenue:

15a + 10s + 5c = 1050

Twice as many child tickets as adult tickets:

c = 2a

Substitute c = 2a into the total ticket equation:

3a + s = 100

Therefore:

s = 100 − 3a

Substitute into the revenue equation:

15a + 10(100 − 3a) + 5(2a) = 1050

15a + 1000 − 30a + 10a = 1050

−5a = 50

a = −10

Then:

c = −20

s = 100 − 3(−10) = 130

The algebraic solution is:

a = −10

s = 130

c = −20

Verification:

−10 + 130 − 20 = 100

15(−10) + 10(130) + 5(−20)

= −150 + 1300 − 100

= 1050

And:

−20 = 2(−10)

The values satisfy the equations, but negative ticket quantities are impossible.

Answer: The system has a unique algebraic solution but no valid real-world solution using nonnegative ticket quantities.

Final Summary

Elimination provides a systematic method for solving three-variable systems of linear equations.

By combining equations to eliminate the same variable twice, we reduce the original system to two equations in two unknowns. We can then solve the smaller system and use back-substitution to determine the remaining variable.

Clear organisation, careful handling of signs and coefficients, and verification are essential for avoiding errors.

The most important principle is to simplify the system one variable at a time while preserving the relationships represented by the original equations.

3. Applications of Three Variables

Introduction

Many real-world problems involve three unknown quantities that are connected by several relationships. These problems appear in shopping, business, transport, science, engineering, agriculture, and everyday decision-making.

For example, a school might sell three types of tickets, a business might manufacture three different products, or a scientist might combine three substances to create a mixture.

When three unknown quantities must be determined simultaneously, we can often represent the situation using a system of three linear equations.

By solving the system, we can determine values that satisfy all the conditions of the problem.

However, solving the equations is only part of the process. We must also interpret the answers, include appropriate units, and check that the results make sense in the original situation.

Why Three Variables Are Useful

Consider a school shop that sells notebooks, pens, and pencils.

Suppose we want to determine the price of each item.

We have three unknown prices:

• Price of one notebook.

• Price of one pen.

• Price of one pencil.

We can represent these using:

n = notebook price

p = pen price

c = pencil price

If customers purchase different combinations of these items, each purchase can provide an equation.

With three suitable independent equations, we can often determine all three prices.

A simple example

Suppose:

• One notebook, one pen, and one pencil cost $9.

• Two notebooks, one pen, and one pencil cost $13.

• One notebook, two pens, and one pencil cost $12.

The equations are:

n + p + c = 9

2n + p + c = 13

n + 2p + c = 12

Subtract the first equation from the second:

n = 4

Subtract the first equation from the third:

p = 3

Substitute into the first equation:

4 + 3 + c = 9

c = 2

Therefore:

Notebook = $4, pen = $3, pencil = $2.

Verification:

4 + 3 + 2 = 9

2(4) + 3 + 2 = 13

4 + 2(3) + 2 = 12

All three purchases are represented correctly.

This demonstrates how information from several situations can be combined to determine unknown values.

The Mathematical Structure of Three-Variable Applications

A linear equation in three variables has the general form:

ax + by + cz = d

where:

• x, y, and z represent unknown quantities.

• a, b, and c are coefficients.

• d is a constant.

A three-variable system contains several equations involving the same unknowns.

For example:

2x + 3y + z = 20

x + 2y + 4z = 19

3x + y + 2z = 23

The equations may represent quantities, costs, distances, masses, or other relationships.

A solution is an ordered triple:

(x, y, z)

that satisfies every equation.

Conditions for a meaningful solution

A successful application solution must satisfy two requirements.

Mathematical correctness: The values satisfy all the equations.

Contextual correctness: The values make sense in the real-world situation.

For example, a negative number of tickets is mathematically possible as an algebraic result, but it cannot represent an actual ticket sale.

Similarly, 12.5 students is not a valid count of individual students.

Translating Word Problems into Equations

The most important part of an application problem is often constructing the equations correctly.

Consider these common phrases:

Notice that some relationships involve only two variables. They can still form part of a three-variable system.

A General Strategy for Application Problems

Step 1: Identify the unknown quantities

Determine exactly what must be calculated.

Step 2: Define variables

Use clear definitions and include units.

For example:

x = number of adult tickets

y = number of student tickets

z = number of child tickets

Step 3: Organise the information

Use a table when several categories or measurements are involved.

Step 4: Construct equations

Translate each independent piece of information into an equation.

Step 5: Solve the system

Use elimination, substitution, or a combination of both.

Step 6: Interpret and verify

Explain the answer in words, check the original equations, and evaluate whether the values are realistic.

Application: Finding the Prices of Three Products

A school stationery shop sells notebooks, pens, and pencils.

Three customers make the following purchases:

• Customer A buys 2 notebooks, 1 pen, and 3 pencils for $16.

• Customer B buys 1 notebook, 3 pens, and 2 pencils for $15.

• Customer C buys 3 notebooks, 2 pens, and 1 pencil for $23.

Find the price of each item.

Step 1: Define variables

n = price of one notebook in dollars

p = price of one pen in dollars

c = price of one pencil in dollars

Step 2: Organise the information

Step 3: Construct equations

A: 2n + p + 3c = 16

B: n + 3p + 2c = 15

C: 3n + 2p + c = 23

Step 4: Eliminate n

Multiply B by 2:

2n + 6p + 4c = 30

Subtract A:

5p + c = 14

Call this D.

Multiply B by 3:

3n + 9p + 6c = 45

Subtract C:

7p + 5c = 22

Call this E.

Step 5: Solve D and E

D: 5p + c = 14

E: 7p + 5c = 22

From D:

c = 14 − 5p

Substitute into E:

7p + 5(14 − 5p) = 22

7p + 70 − 25p = 22

−18p = −48

p = 8/3

Then:

c = 14 − 5(8/3)

c = 2/3

Step 6: Find n

Use B:

n + 3p + 2c = 15

n + 3(8/3) + 2(2/3) = 15

n + 8 + 4/3 = 15

n = 17/3

Step 7: Interpret

Notebook price = $17/3, approximately $5.67

Pen price = $8/3, approximately $2.67

Pencil price = $2/3, approximately $0.67

These are exact mathematical prices; the rounded amounts should not be substituted back as if they were exact cent-based prices.

Step 8: Verify using exact values

A:

2(17/3) + 8/3 + 3(2/3)

= 34/3 + 8/3 + 6/3

= 48/3 = 16

B:

17/3 + 3(8/3) + 2(2/3)

= 17/3 + 24/3 + 4/3

= 45/3 = 15

C:

3(17/3) + 2(8/3) + 2/3

= 51/3 + 16/3 + 2/3

= 69/3 = 23

Answer: Notebook = $17/3, pen = $8/3, pencil = $2/3.

The mathematical solution is valid, although a real shop pricing items to the nearest cent would need additional pricing constraints.

Application: Ticket Sales at a School Event

A school sells three types of tickets for a concert.

• Adult tickets cost $15.

• Student tickets cost $10.

• Child tickets cost $5.

A total of 150 tickets are sold, generating $1,500 in revenue.

The number of student tickets is twice the number of child tickets.

Find how many tickets of each type were sold.

Step 1: Define variables

a = number of adult tickets

s = number of student tickets

c = number of child tickets

Step 2: Construct equations

Total tickets:

a + s + c = 150

Total revenue:

15a + 10s + 5c = 1500

Relationship between student and child tickets:

s = 2c

Step 3: Substitute the relationship

Substitute s = 2c into the first equation:

a + 2c + c = 150

a + 3c = 150

Therefore:

a = 150 − 3c

Step 4: Substitute into the revenue equation

15(150 − 3c) + 10(2c) + 5c = 1500

2250 − 45c + 20c + 5c = 1500

2250 − 20c = 1500

−20c = −750

c = 37.5

Then:

s = 75

a = 37.5

Step 5: Interpret

The system has a unique algebraic solution:

a = 37.5

s = 75

c = 37.5

However, ticket quantities must be whole numbers.

Therefore, the information does not produce a valid real-world solution.

This is an important distinction between solving an equation and solving a practical problem.

Adjusting the information

Suppose the total revenue had been $1,450 instead.

Then:

2250 − 20c = 1450

20c = 800

c = 40

s = 80

a = 150 − 120 = 30

Verification:

30 + 80 + 40 = 150 tickets

15(30) + 10(80) + 5(40)

= 450 + 800 + 200

= $1450

With the revised revenue, the school sold 30 adult tickets, 80 student tickets, and 40 child tickets.

Application: Mixing Three Solutions

Systems of three variables are especially useful in chemistry and manufacturing.

Suppose a laboratory needs to prepare 100 litres of a solution with a concentration of 30%.

Three available solutions have concentrations of:

• 10%

• 30%

• 50%

The laboratory must use twice as much 10% solution as 50% solution.

How many litres of each solution should be used?

Step 1: Define variables

x = volume of 10% solution, in litres

y = volume of 30% solution, in litres

z = volume of 50% solution, in litres

Step 2: Construct the volume equation

The total volume is 100 litres.

x + y + z = 100

Step 3: Construct the concentration equation

The amount of dissolved substance in each solution is:

10% solution: 0.10x

30% solution: 0.30y

50% solution: 0.50z

The final mixture contains:

0.30 × 100 = 30 litres of dissolved substance

Therefore:

0.10x + 0.30y + 0.50z = 30

This model assumes volumes are additive.

Step 4: Construct the relationship equation

The volume of 10% solution is twice the volume of 50% solution:

x = 2z

Step 5: Simplify the concentration equation

Multiply by 10:

x + 3y + 5z = 300

Subtract the total volume equation:

(x + 3y + 5z) − (x + y + z) = 300 − 100

2y + 4z = 200

Divide by 2:

y + 2z = 100

Step 6: Use the relationship equation

Substitute x = 2z into the volume equation:

2z + y + z = 100

y + 3z = 100

We now have:

y + 2z = 100

y + 3z = 100

Subtract:

z = 0

Then:

x = 0

y = 100

Step 7: Interpret

The mathematical solution is:

• 10% solution: 0 litres

• 30% solution: 100 litres

• 50% solution: 0 litres

This is mathematically valid, but it does not involve combining all three solutions.

If the laboratory must use a positive amount of every solution, the conditions are impossible.

Why does this happen?

The final concentration is 30%, which is exactly the concentration of the middle solution.

To balance 10% and 50% solutions around 30%, equal volumes of the two would be needed.

However, the problem requires twice as much 10% solution as 50% solution.

These conditions can only be satisfied when neither of those two solutions is used.

This demonstrates how algebra can reveal limitations in a proposed scientific procedure.

Application: A Three-Ingredient Mixture

A food manufacturer produces a 100 kg mixture using three ingredients.

Ingredient A costs $2 per kilogram.

Ingredient B costs $4 per kilogram.

Ingredient C costs $6 per kilogram.

The mixture costs $380 in total.

The manufacturer uses 10 kg more of Ingredient A than Ingredient C.

Determine the mass of each ingredient.

Step 1: Define variables

x = mass of Ingredient A, in kg

y = mass of Ingredient B, in kg

z = mass of Ingredient C, in kg

Step 2: Construct equations

Total mass:

x + y + z = 100

Total cost:

2x + 4y + 6z = 380

Relationship:

x − z = 10

Step 3: Simplify the cost equation

Divide by 2:

x + 2y + 3z = 190

Subtract the mass equation:

y + 2z = 90

Step 4: Use the relationship equation

x = z + 10

Substitute into the mass equation:

(z + 10) + y + z = 100

y + 2z = 90

This is the same equation obtained from the cost information.

Step 5: Interpret the result

We have:

x = z + 10

y = 90 − 2z

The solutions are:

(x, y, z) = (z + 10, 90 − 2z, z)

For nonnegative ingredient masses:

0 ≤ z ≤ 45

Therefore, there are infinitely many possible mixtures.

For example:

All three mixtures have a mass of 100 kg and cost $380.

What does this teach us?

Three equations do not necessarily provide three independent pieces of information.

In this example, one equation is redundant.

An additional independent condition is needed to determine a unique mixture.

Application: Three Vehicles and Travel Distances

A delivery company uses three vehicles.

Vehicle A travels at 40 km/h.

Vehicle B travels at 60 km/h.

Vehicle C travels at 80 km/h.

Each vehicle travels for a different amount of time.

Together, they travel for 12 hours and cover 720 km.

Vehicle A travels for twice as long as Vehicle C.

Determine how long each vehicle travels.

Step 1: Define variables

x = travel time of Vehicle A, in hours

y = travel time of Vehicle B, in hours

z = travel time of Vehicle C, in hours

Step 2: Construct equations

Total travel time:

x + y + z = 12

Total distance:

40x + 60y + 80z = 720

Relationship:

x = 2z

Step 3: Simplify the distance equation

Divide by 20:

2x + 3y + 4z = 36

Multiply the time equation by 2:

2x + 2y + 2z = 24

Subtract:

y + 2z = 12

Step 4: Substitute the relationship

From x = 2z:

2z + y + z = 12

y + 3z = 12

Subtract the reduced equations:

z = 0

Then:

x = 0

y = 12

Step 5: Interpret

The only nonnegative solution is:

Vehicle A: 0 hours

Vehicle B: 12 hours

Vehicle C: 0 hours

This result satisfies the algebra but is unsuitable if all three vehicles were required to operate.

It also makes sense physically: the required average speed is 720/12 = 60 km/h. The stated relationship between the slower and faster vehicles prevents their contributions from balancing unless both are zero.

Improving the model

If the company instead specifies that Vehicle A travels for the same amount of time as Vehicle C:

x = z

Then:

x + y + z = 12

2x + 3y + 4z = 36

x = z

Substitute x = z:

2z + y = 12

6z + 3y = 36

Dividing the second equation by 3 gives the same relationship.

This modified system also has infinitely many solutions.

The important lesson is that the equations must provide independent information to determine a unique result.

Application: Production Planning in a Factory

A factory manufactures three types of products: A, B, and C.

During one production period:

• The factory manufactures 200 products in total.

• Product A requires 2 machine-hours, Product B requires 3 machine-hours, and Product C requires 5 machine-hours.

• The total machine time used is 650 hours.

• The factory manufactures 20 more units of Product A than Product C.

Find the number of each product manufactured.

Step 1: Define variables

x = number of Product A units

y = number of Product B units

z = number of Product C units

Step 2: Construct equations

Total products:

x + y + z = 200

Total machine-hours:

2x + 3y + 5z = 650

Quantity relationship:

x − z = 20

Step 3: Eliminate x

Multiply the first equation by 2:

2x + 2y + 2z = 400

Subtract from the machine-hours equation:

y + 3z = 250

Step 4: Use the third equation

x = z + 20

Substitute into the total products equation:

z + 20 + y + z = 200

y + 2z = 180

Step 5: Solve

Subtract:

(y + 3z) − (y + 2z) = 250 − 180

z = 70

Then:

y + 2(70) = 180

y = 40

And:

x = 70 + 20

x = 90

Step 6: Verify

Total products:

90 + 40 + 70 = 200

Machine-hours:

2(90) + 3(40) + 5(70)

= 180 + 120 + 350

= 650

Quantity difference:

90 − 70 = 20

Answer: The factory manufactures 90 units of A, 40 units of B, and 70 units of C.

This is a practical example of using three variables to make production and resource-planning decisions.

Application: Three Types of Transport Tickets

A transport company sells single, day, and weekly passes.

• A single ticket costs $3.

• A day pass costs $8.

• A weekly pass costs $20.

During a sales period, the company sells 100 passes and tickets, earning $800.

It sells 20 more single tickets than day passes.

Determine how many of each type are sold.

Step 1: Define variables

x = number of single tickets

y = number of day passes

z = number of weekly passes

Step 2: Construct equations

Total sales:

x + y + z = 100

Total revenue:

3x + 8y + 20z = 800

Relationship:

x − y = 20

Step 3: Substitute

x = y + 20

From the total sales equation:

(y + 20) + y + z = 100

z = 80 − 2y

Step 4: Solve

Substitute into the revenue equation:

3(y + 20) + 8y + 20(80 − 2y) = 800

3y + 60 + 8y + 1600 − 40y = 800

1660 − 29y = 800

29y = 860

y = 860/29

Then:

x = 1440/29

z = 600/29

Step 5: Interpret

These values are not whole numbers.

Since tickets and passes are sold as complete items, the model does not produce a valid solution for actual sales.

A business could use this finding to identify inconsistent records or incorrect assumptions.

Application: A Three-Subject Examination

A student completes three assessments in mathematics, science, and English.

The total of the three scores is 240 points.

The mathematics score is 10 points higher than the science score.

The English score is 20 points lower than the mathematics score.

Find the student's score in each subject.

Step 1: Define variables

m = mathematics score

s = science score

e = English score

Step 2: Construct equations

Total:

m + s + e = 240

Mathematics compared with science:

m − s = 10

English compared with mathematics:

m − e = 20

Step 3: Express variables in terms of m

s = m − 10

e = m − 20

Step 4: Substitute

m + (m − 10) + (m − 20) = 240

3m − 30 = 240

3m = 270

m = 90

Then:

s = 80

e = 70

Step 5: Verify

90 + 80 + 70 = 240

90 − 80 = 10

90 − 70 = 20

Answer: Mathematics = 90 points, Science = 80 points, English = 70 points.

This example demonstrates that not every three-variable application requires complicated elimination. Substitution may be more efficient when two relationships already express variables in terms of another.

Application: Agricultural Planning

A farmer plants three crops: corn, wheat, and vegetables.

The farmer has 50 hectares available.

• Corn requires 4 labour-hours per hectare.

• Wheat requires 2 labour-hours per hectare.

• Vegetables require 6 labour-hours per hectare.

The farmer has 190 labour-hours available.

The area planted with corn is 5 hectares greater than the area planted with vegetables.

Determine the area used for each crop.

Step 1: Define variables

c = hectares planted with corn

w = hectares planted with wheat

v = hectares planted with vegetables

Step 2: Construct equations

Total land:

c + w + v = 50

Total labour:

4c + 2w + 6v = 190

Relationship:

c − v = 5

Step 3: Eliminate w

Multiply the land equation by 2:

2c + 2w + 2v = 100

Subtract from the labour equation:

2c + 4v = 90

Divide by 2:

c + 2v = 45

Step 4: Use the relationship equation

c = v + 5

Substitute:

(v + 5) + 2v = 45

3v + 5 = 45

3v = 40

v = 40/3

Step 5: Find the other variables

c = 40/3 + 5

c = 55/3

Then:

w = 50 − 55/3 − 40/3

w = 55/3

Step 6: Interpret

Corn: 55/3 hectares, approximately 18.33 ha

Wheat: 55/3 hectares, approximately 18.33 ha

Vegetables: 40/3 hectares, approximately 13.33 ha

Step 7: Verify

Total land:

55/3 + 55/3 + 40/3 = 150/3 = 50 ha

Total labour:

4(55/3) + 2(55/3) + 6(40/3)

= 220/3 + 110/3 + 240/3

= 570/3 = 190 hours

Difference:

55/3 − 40/3 = 15/3 = 5 ha

Answer: Corn = 55/3 ha, wheat = 55/3 ha, vegetables = 40/3 ha.

Unlike counts of people or tickets, fractional hectares are reasonable because land area is a continuous quantity.

How to Explain Mathematical Reasoning

A complete solution should communicate the reasoning behind the equations, not just the calculations.

Consider a school that sells three types of meals.

A complete explanation should identify:

What the variables represent

For example:

x = number of vegetarian meals

y = number of chicken meals

z = number of fish meals

Why each equation is appropriate

If 120 meals were sold:

x + y + z = 120

If the meals cost $6, $8, and $10 respectively, and total revenue is $920:

6x + 8y + 10z = 920

If 10 more vegetarian meals than fish meals were sold:

x − z = 10

How the system is solved

Explain whether elimination or substitution is being used, and show the important intermediate steps.

What the solution means

Write the number of each type of meal in words.

Why the solution is reasonable

Check that the quantities are nonnegative whole numbers and that the totals match the original information.

This approach makes mathematical work easier to follow and evaluate.

Comparing Different Application Types

The same algebraic methods can be applied across many different contexts.

Recognising When a Model Is Incomplete

Sometimes a problem does not contain enough independent information to determine all three unknowns.

For example:

x + y + z = 100

2x + 2y + 2z = 200

x − y = 10

The second equation is twice the first.

It does not provide any new information.

Therefore, the system has infinitely many real-number solutions.

Why this matters in real life

If a business uses incomplete information, it may be unable to determine exact quantities.

If a scientist has insufficient experimental measurements, several combinations of substances may fit the data.

If a school has incomplete sales records, it may be impossible to determine the number of each ticket type sold.

Mathematics helps identify when additional information is needed.

Recognising Inconsistent Information

Consider:

x + y + z = 100

x + y + z = 120

The equations contradict each other.

The same three quantities cannot have two different totals at the same time.

Therefore, no solution exists.

In a real-world situation, this might indicate:

• A recording error.

• An incorrect assumption.

• Inconsistent measurements.

• Missing or inaccurate information.

Recognising inconsistencies is an important practical application of systems of equations.

Common Mistakes

Defining variables unclearly

Always explain what each variable represents and include units where appropriate.

Constructing the wrong equation

Read statements carefully. "Twice as many" means multiplication, while "two more" means addition.

Confusing quantity and cost

The number of items and their prices are different quantities.

Ignoring units

Mass, volume, distance, time, and money must be handled consistently.

Assuming three equations always produce one solution

Dependent or inconsistent equations may prevent a unique solution.

Stopping after finding two variables

A three-variable problem requires values for all three unknowns unless the system has no unique solution.

Accepting impossible values

Negative quantities of products or fractional numbers of people may be invalid.

Rounding too early

Keep exact fractions during calculations, especially when verifying answers.

Failing to explain the meaning of the result

A final answer should state what each value represents.

Not checking the original conditions

Substitute values into all original equations and verify contextual restrictions.

Key Terms

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

• Three-variable system: A system of equations involving three unknown quantities.

• Variable: A symbol representing an unknown quantity.

• Coefficient: A number multiplying a variable.

• Linear equation: An equation in which variables occur only to the first power and are not multiplied together.

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

• Ordered triple: Three values written in a specified order.

• Elimination: Removing a variable by combining equations.

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

• Independent equations: Equations providing distinct mathematical information.

• Dependent equations: Equations containing redundant information.

• Consistent system: A system with at least one solution.

• Inconsistent system: A system with no solution.

• Constraint: A restriction on allowable values.

• Revenue: Money received from sales.

• Unit price: The cost of one unit of a product.

• Concentration: The amount of a substance relative to the amount of solution.

• Resource allocation: Distributing available resources among different uses.

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

• Interpretation: Explaining what a mathematical result means in context.

Key Takeaways

• Three-variable systems can represent many practical situations involving three unknown quantities.

• Variables must be defined clearly before equations are constructed.

• Tables help organise complex information.

• Each equation should represent a specific relationship from the problem.

• Elimination and substitution can be used to solve three-variable systems.

• A unique solution requires sufficient independent information.

• Some systems have no solution or infinitely many solutions.

• Mathematical solutions must be interpreted using appropriate units.

• Contextual restrictions may exclude algebraically valid answers.

• Verification requires checking every original condition.

• Clear explanations connect mathematical calculations to real-world meaning.

Check Your Understanding

Recall and Understanding

1. Why are three variables useful in real-world mathematical problems?

2. Explain why variables should be defined before constructing equations.

3. What is the difference between a mathematically correct solution and a contextually valid solution?

4. Why might a system of three equations fail to produce a unique solution?

5. Explain how tables can help when constructing equations from word problems.

Application

6. A shop sells notebooks, pens, and pencils.

• One notebook, one pen, and one pencil cost $10.

• Two notebooks, one pen, and one pencil cost $15.

• One notebook, two pens, and one pencil cost $13.

Find the price of each item.

7. A theatre sells adult tickets for $15, student tickets for $10, and child tickets for $5.

A total of 100 tickets are sold, generating $1,000.

There are twice as many student tickets as child tickets.

Find the number of each ticket type.

8. A company manufactures three products.

Product A requires 2 machine-hours per unit.

Product B requires 3 machine-hours per unit.

Product C requires 4 machine-hours per unit.

The company manufactures 100 units using 280 machine-hours.

It produces 10 more units of A than C.

Determine the number of each product.

9. Three numbers add to 90.

The first number is twice the second.

The third number is 15 greater than the second.

Find the numbers.

10. A student scores a total of 255 points across three tests.

The mathematics score is 10 points higher than the science score.

The English score is 5 points lower than the science score.

Find the three scores.

Analysis and Challenge

11. A laboratory combines three solutions with concentrations of 10%, 20%, and 40%.

The total volume is 100 litres, and the final concentration is 25%.

The laboratory uses equal volumes of the 10% and 40% solutions.

Find the volume of each solution.

12. A farmer plants corn, wheat, and vegetables across 60 hectares.

Corn requires 4 labour-hours per hectare, wheat requires 2, and vegetables require 6.

The total labour requirement is 240 hours.

The corn area is twice the wheat area.

Determine the area of each crop.

13. A school sells three types of lunch.

Vegetarian meals cost $6, chicken meals cost $8, and fish meals cost $10.

A total of 120 meals are sold, generating $960.

The number of vegetarian meals equals the number of fish meals.

Determine whether the system has a unique solution.

14. A company records:

x + y + z = 150

2x + 2y + 2z = 320

x − y = 10

Explain why the system has no solution.

15. A transport company sells single tickets for $3, day passes for $8, and weekly passes for $20.

A total of 100 tickets and passes are sold, generating $800.

There are 20 more single tickets than day passes.

Construct the system, solve it algebraically, and explain whether the solution is valid in context.

Answers to Check Your Understanding

1. Three variables allow us to represent three unknown quantities simultaneously, such as three prices, three product quantities, or three ingredient amounts.

2. Defining variables ensures that each equation represents the correct quantities and allows the final answer to be interpreted clearly.

3. A mathematically correct solution satisfies the equations. A contextually valid solution also satisfies real-world restrictions such as nonnegative quantities, whole-number counts, and appropriate units.

4. The equations may be dependent or inconsistent. Three equations do not guarantee three independent pieces of information.

5. Tables organise quantities, coefficients, and totals, making it easier to translate information accurately into equations.

6.

Let n = notebook price, p = pen price, and c = pencil price.

n + p + c = 10

2n + p + c = 15

n + 2p + c = 13

Subtract the first equation from the second:

n = 5

Subtract the first from the third:

p = 3

Substitute:

5 + 3 + c = 10

c = 2

Answer: Notebook = $5, pen = $3, pencil = $2.

Verification:

5 + 3 + 2 = 10

2(5) + 3 + 2 = 15

5 + 2(3) + 2 = 13

7.

Let a = adult tickets, s = student tickets, and c = child tickets.

a + s + c = 100

15a + 10s + 5c = 1000

s = 2c

Substitute s = 2c:

a + 3c = 100

a = 100 − 3c

Substitute into revenue:

15(100 − 3c) + 10(2c) + 5c = 1000

1500 − 45c + 20c + 5c = 1000

1500 − 20c = 1000

c = 25

Then:

s = 50

a = 25

Answer: 25 adult tickets, 50 student tickets, and 25 child tickets.

Verification:

25 + 50 + 25 = 100

15(25) + 10(50) + 5(25) = 1000

8.

Let x = Product A, y = Product B, and z = Product C.

x + y + z = 100

2x + 3y + 4z = 280

x − z = 10

Multiply the first equation by 2 and subtract:

y + 2z = 80

From the third equation:

x = z + 10

Substitute into the total quantity equation:

y + 2z = 90

The reduced equations contradict each other:

y + 2z = 80

y + 2z = 90

Answer: No solution. The production information is inconsistent.

9.

Let x, y, and z represent the three numbers.

x + y + z = 90

x = 2y

z = y + 15

Substitute:

2y + y + y + 15 = 90

4y = 75

y = 18.75

Then:

x = 37.5

z = 33.75

Answer: 37.5, 18.75, and 33.75.

Verification:

37.5 + 18.75 + 33.75 = 90

10.

Let m = mathematics, s = science, and e = English.

m + s + e = 255

m = s + 10

e = s − 5

Substitute:

(s + 10) + s + (s − 5) = 255

3s + 5 = 255

3s = 250

s = 250/3

Then:

m = 280/3

e = 235/3

Answer: Mathematics = 93⅓ points, Science = 83⅓ points, English = 78⅓ points.

These satisfy the equations. If the tests award only whole points, the given conditions would not have a valid whole-number solution.

11.

Let x, y, and z be the volumes of 10%, 20%, and 40% solutions.

x + y + z = 100

0.10x + 0.20y + 0.40z = 25

x = z

Substitute x = z into the volume equation:

y + 2z = 100

Therefore:

y = 100 − 2z

Substitute into the concentration equation:

0.10z + 0.20(100 − 2z) + 0.40z = 25

0.10z + 20 − 0.40z + 0.40z = 25

0.10z = 5

z = 50

Then:

x = 50

y = 0

Answer: 50 L of 10% solution, 0 L of 20% solution, and 50 L of 40% solution.

Verification:

Total volume = 100 L

Dissolved substance = 5 + 20 = 25 L

Concentration = 25%

12.

Let c = corn area, w = wheat area, and v = vegetable area.

c + w + v = 60

4c + 2w + 6v = 240

c = 2w

Substitute:

2w + w + v = 60

v = 60 − 3w

Into the labour equation:

4(2w) + 2w + 6(60 − 3w) = 240

8w + 2w + 360 − 18w = 240

−8w = −120

w = 15

Then:

c = 30

v = 15

Answer: Corn = 30 ha, wheat = 15 ha, vegetables = 15 ha.

Verification:

30 + 15 + 15 = 60

4(30) + 2(15) + 6(15) = 240

13.

Let v = vegetarian meals, c = chicken meals, and f = fish meals.

v + c + f = 120

6v + 8c + 10f = 960

v = f

Substitute f = v:

2v + c = 120

The revenue equation becomes:

16v + 8c = 960

Divide by 8:

2v + c = 120

Both equations provide the same information.

Therefore, the system does not have a unique solution.

Let v = t.

Then:

f = t

c = 120 − 2t

Answer: Infinitely many nonnegative real solutions, with 0 ≤ t ≤ 60. For whole meal counts, t must be an integer, giving 61 possible combinations.

14.

The first equation states:

x + y + z = 150

Multiplying by 2 gives:

2x + 2y + 2z = 300

But the second equation states:

2x + 2y + 2z = 320

This is a contradiction.

Answer: No solution. The information is inconsistent.

15.

Let x = single tickets, y = day passes, and z = weekly passes.

x + y + z = 100

3x + 8y + 20z = 800

x − y = 20

From the third equation:

x = y + 20

Substitute into the total quantity equation:

z = 80 − 2y

Substitute into revenue:

3(y + 20) + 8y + 20(80 − 2y) = 800

3y + 60 + 8y + 1600 − 40y = 800

−29y = −860

y = 860/29

Then:

x = 1440/29

z = 600/29

The exact algebraic solution is:

x = 1440/29

y = 860/29

z = 600/29

These values are not integers.

Answer: The system has a unique algebraic solution but no valid whole-number ticket solution.

Final Summary

Three-variable systems are powerful tools for representing real-world situations involving three unknown quantities.

They can be applied to prices, ticket sales, mixtures, manufacturing, transport, assessment scores, agriculture, and many other practical problems.

The essential process is to identify the unknown quantities, construct accurate equations, solve the system, and interpret the results using the original context.

Not every practical problem produces a unique or realistic solution. Sometimes the information is incomplete, inconsistent, or incompatible with real-world restrictions.

A successful mathematical model must do more than satisfy equations. It must also provide an explanation that makes sense in the situation being studied.

4. Technology for Solving Systems

Introduction

Systems of equations are used to solve problems involving several unknown quantities. They appear in mathematics, science, engineering, economics, business, and many other fields.

We have already learned how to solve systems using algebraic methods such as substitution and elimination. We have also explored how systems can be represented graphically, with solutions appearing where lines or planes intersect.

Although these manual methods are important, some systems become difficult to solve because they involve complicated coefficients, fractions, decimals, or several variables.

This is where technology becomes especially useful.

Graphing calculators, online graphing tools, spreadsheets, and computer algebra systems can help us:

• Graph equations and identify intersection points.

• Solve systems of two or more variables.

• Check answers obtained using manual methods.

• Investigate systems with no solution or infinitely many solutions.

• Interpret mathematical results in real-world situations.

• Explore how changing an equation affects its solution.

However, technology does not replace mathematical understanding. A calculator may produce an answer, but we still need to understand what the answer represents, whether it is accurate, and whether it makes sense.

Why Use Technology to Solve Systems?

Consider the system:

2x + 3y = 12

x − y = 1

This system can be solved easily by substitution or elimination.

However, consider:

17x + 23y = 146

31x − 19y = 52

The second system is more time-consuming to solve manually.

Technology can help determine the solution quickly and reduce arithmetic errors.

Advantages of technology

Speed: Complex calculations can be completed quickly.

Accuracy: Software can reduce arithmetic mistakes, provided equations are entered correctly.

Visualisation: Graphs help us understand how equations relate to one another.

Verification: We can check manually calculated solutions.

Exploration: We can change coefficients and observe how solutions change.

Larger systems: Computers can solve systems involving many unknowns.

Limitations of technology

Technology also has limitations.

• Incorrectly entered equations produce incorrect results.

• Graphical solutions may be approximate.

• Some systems cannot be solved uniquely.

• Numerical methods may introduce rounding errors.

• Software may display results without explaining the reasoning.

• A mathematically correct answer may be unrealistic in context.

The best approach is to combine mathematical reasoning with technological tools.

Graphing Systems of Two Linear Equations

A system of two linear equations can be represented by two lines on the same coordinate plane.

For example:

y = 2x + 1

y = −x + 7

The solution is the point where the lines intersect.

Step 1: Enter the equations

In a graphing calculator or graphing application, enter:

y = 2x + 1

y = −x + 7

Step 2: Display both graphs

The software draws both lines on the same coordinate plane.

Step 3: Locate the intersection

Use the graphing tool's intersection feature, or zoom in and inspect the crossing point.

The intersection is:

(2, 5)

Step 4: Interpret

The x-coordinate is 2.

The y-coordinate is 5.

Therefore:

x = 2

y = 5

Step 5: Verify algebraically

First equation:

5 = 2(2) + 1

5 = 5

Second equation:

5 = −2 + 7

5 = 5

Both equations are satisfied.

Final answer: (2, 5).

Using Desmos to Solve Systems

Desmos Graphing Calculator is a widely used graphing tool that allows users to enter equations and explore their graphs.

You can access it at Desmos Graphing Calculator.

Example

Solve:

3x + 2y = 12

x − y = 1

Step 1: Enter both equations

In Desmos, type:

3x + 2y = 12

x − y = 1

Desmos can graph equations written in standard form, so rearranging them into y = mx + b is not always necessary.

Step 2: Identify the intersection

The lines intersect at:

(2.8, 1.8)

Step 3: Verify

First equation:

3(2.8) + 2(1.8)

= 8.4 + 3.6

= 12

Second equation:

2.8 − 1.8 = 1

Answer: x = 2.8 and y = 1.8.

Why graphing is useful

The graph shows that the two lines cross at exactly one point.

This helps us understand that the system has a unique solution.

Graphing is particularly useful for visualising systems and checking answers.

Understanding Graphing Windows

A graphing window determines which part of the coordinate plane is visible.

For example, suppose:

y = 2x + 100

y = −x + 400

The intersection occurs at:

2x + 100 = −x + 400

3x = 300

x = 100

y = 300

The solution is:

(100, 300)

If the graphing window only displays x-values between −10 and 10 and y-values between −10 and 10, the intersection will not be visible.

This does not mean the system has no solution.

It means the graphing window is too small.

Improving the graph

Adjust the axes to include the expected solution.

For example:

x-axis: 0 to 150

y-axis: 0 to 450

The intersection will then be visible.

Important lesson

A missing intersection on the screen does not necessarily mean that the system has no solution.

Always consider the scale and viewing window.

Using GeoGebra to Solve Systems

GeoGebra is another useful tool for exploring equations and their solutions.

GeoGebra Graphing Calculator

GeoGebra can graph equations, identify intersections, and explore how changing coefficients affects a system.

Worked Example: Two Lines

Solve:

2x + y = 11

x − y = 1

Enter both equations into the graphing calculator.

The two lines intersect at:

(4, 3)

Verification:

2(4) + 3 = 11

4 − 3 = 1

Both equations are satisfied.

Comparing graphing tools

For three-variable systems, use GeoGebra 3D Calculator rather than the ordinary two-dimensional graphing view.

Graphing Systems with No Solution

Consider:

y = 2x + 1

y = 2x − 4

Both equations have the same slope:

m = 2

However, they have different y-intercepts:

1 and −4

Therefore, the lines are parallel.

They never intersect.

A graphing calculator will display two distinct parallel lines.

Answer: No solution.

Verifying algebraically

Set the expressions equal:

2x + 1 = 2x − 4

Subtract 2x:

1 = −4

This is impossible.

The system is inconsistent.

Graphing Systems with Infinitely Many Solutions

Consider:

y = 3x + 2

2y = 6x + 4

Simplify the second equation:

y = 3x + 2

Both equations represent exactly the same line.

The graphs overlap completely.

Every point on the line satisfies both equations.

Answer: Infinitely many solutions.

Interpreting overlapping graphs

Graphing software may display what appears to be only one line.

This does not necessarily mean that one equation failed to graph.

It may mean that both equations represent the same line.

To confirm, simplify both equations algebraically.

Comparing the Three Types of Systems

This classification is particularly useful when interpreting technological outputs.

Solving Three-Variable Systems with Technology

Two-variable systems can be represented by lines in a coordinate plane.

Three-variable linear systems are different.

Each equation generally represents a plane in three-dimensional space.

For example:

x + y + z = 9

2x + y + z = 11

x + 2y + z = 12

Each equation represents a plane.

The solution is the point shared by all three planes.

Worked Example: Three-Variable System

Solve:

x + y + z = 9

2x + y + z = 11

x + 2y + z = 12

Using graphing technology

Enter the equations into a suitable three-dimensional graphing application.

The software displays three planes.

If the planes share exactly one point, the system has a unique solution.

For this system, the common point is:

(2, 3, 4)

Verify manually

First equation:

2 + 3 + 4 = 9

Second equation:

2(2) + 3 + 4 = 11

Third equation:

2 + 2(3) + 4 = 12

All three equations are satisfied.

Confirm using elimination

Subtract the first equation from the second:

x = 2

Subtract the first equation from the third:

y = 3

Substitute into the first:

2 + 3 + z = 9

z = 4

Answer: (2, 3, 4).

The graphical and algebraic methods produce the same solution.

Interpreting Three-Dimensional Graphs

Three-dimensional graphs require careful interpretation because the planes may overlap or intersect in different ways.

One solution

Three planes share exactly one point.

The solution is one ordered triple.

No solution

The three planes have no common point.

For example, two of the planes may be parallel and distinct.

Infinitely many solutions

The planes may share a line or coincide.

In these situations, infinitely many points satisfy the equations.

Important limitation

A three-dimensional picture can be misleading because of perspective, overlapping surfaces, or the selected viewing angle.

A graph may suggest a particular intersection, but algebraic verification is needed to confirm the result.

Using Scientific Calculators

Many scientific calculators contain a simultaneous-equation solver.

For example, some models from Casio and Texas Instruments support solving linear systems.

The exact procedure depends on the calculator model.

A typical workflow is:

1. Select the equation or simultaneous-equation mode.

2. Choose the number of unknowns.

3. Enter the coefficients and constants.

4. Execute the calculation.

5. Read the values of the variables.

6. Substitute the results into the original equations.

Worked Example

Solve:

2x + y − z = 8

x + 2y + z = 7

3x − y + 2z = 7

The coefficients can be organised as follows:

A calculator's simultaneous-equation mode can use these values to solve the system.

The solution is:

x = 3, y = 2, z = 0

Verification:

2(3) + 2 − 0 = 8

3 + 2(2) + 0 = 7

3(3) − 2 + 2(0) = 7

All three equations are satisfied.

Why coefficient tables are useful

Coefficient tables help prevent mistakes when entering equations into a calculator.

They are particularly important when an equation is missing a variable.

For example:

2x + 3z = 10

must be entered as:

2x + 0y + 3z = 10

The coefficient of y is zero.

Forgetting this zero may cause the remaining values to be entered in the wrong positions.

Solving Systems with Spreadsheets

Spreadsheet software such as Microsoft Excel or Google Sheets can also be used to solve systems.

Spreadsheets are especially useful when working with larger sets of equations, repeated calculations, or real-world data.

Representing a system using a matrix

Consider:

2x + y + z = 9

x + 3y + 2z = 13

3x + 2y + z = 11

The coefficients form the matrix:

The constants form another column:

We can represent the system using:

AX = B

where:

A = coefficient matrix

X = column of unknown variables

B = column of constants

Solving with matrix operations

When A is invertible:

X = A⁻¹B

Spreadsheet software can calculate the inverse matrix and multiply it by the constants.

In Excel or Google Sheets, a typical approach is to enter the coefficients in cells A1:C3 and the constants in E1:E3.

Then use:

=MMULT(MINVERSE(A1:C3),E1:E3)

In current spreadsheet versions that support dynamic arrays, the three results can spill into adjacent cells. Older spreadsheet versions may require array-formula entry.

For the example above, the solution is:

x = 2

y = 1

z = 4

Verification:

2(2) + 1 + 4 = 9

2 + 3(1) + 2(4) = 13

3(2) + 2(1) + 4 = 12

The third equation does not equal 11.

Therefore, the proposed values are not the solution.

The correct spreadsheet result is:

x = 3/2, y = −1/2, z = 13/2

Check:

2(3/2) − 1/2 + 13/2 = 9

3/2 + 3(−1/2) + 2(13/2) = 13

3(3/2) + 2(−1/2) + 13/2 = 10

This also fails the third equation.

We therefore need to solve the system carefully rather than accepting an unverified result.

Using elimination:

From the first equation:

z = 9 − 2x − y

Substitute into the second:

x + 3y + 2(9 − 2x − y) = 13

−3x + y = −5

y = 3x − 5

Substitute into the third:

3x + 2y + 9 − 2x − y = 11

x + y = 2

Substitute y = 3x − 5:

x + 3x − 5 = 2

4x = 7

x = 7/4

Then:

y = 3(7/4) − 5

y = 1/4

And:

z = 9 − 2(7/4) − 1/4

z = 21/4

Verification:

2(7/4) + 1/4 + 21/4 = 9

7/4 + 3(1/4) + 2(21/4) = 13

3(7/4) + 2(1/4) + 21/4 = 11

Correct solution: (7/4, 1/4, 21/4).

This example illustrates why verification remains essential, even when technology is involved.

When matrix methods cannot be used directly

The inverse-matrix formula requires an invertible square coefficient matrix.

If the matrix is singular, the inverse does not exist.

This may happen when the equations are dependent or when the system does not have a unique solution.

A calculator or spreadsheet may display an error rather than three numerical values.

That error does not automatically tell us whether the system has no solution or infinitely many solutions.

Further analysis is required.

Comparing Manual and Technological Methods

Different methods have different strengths.

When should technology be used?

Technology is particularly helpful when:

• Coefficients are large or contain fractions.

• A system has three or more unknowns.

• Many similar systems must be solved.

• A graph would improve understanding.

• Manual calculations need checking.

• The effects of changing coefficients are being investigated.

Manual methods are especially useful when the purpose is to understand how the equations work or to explain why a solution exists.

Technology is most effective when it supports mathematical reasoning rather than replacing it.

Worked Example: Comparing Manual and Calculator Solutions

Solve:

4x + 3y = 19

2x − 5y = −13

Manual method

Multiply the second equation by 2:

4x − 10y = −26

Subtract from the first equation:

13y = 45

y = 45/13

Substitute into the first equation:

4x + 3(45/13) = 19

4x = 247/13 − 135/13

4x = 112/13

x = 28/13

Exact solution: (28/13, 45/13).

Calculator method

A calculator may display:

x ≈ 2.153846

y ≈ 3.461538

These are decimal approximations of the same solution.

Comparing the outputs

The manual method gives exact fractions.

The calculator provides convenient decimal values.

Both are useful, but exact fractions are preferable when further algebraic calculations are required.

Technology and Rounding Errors

Computers and calculators often represent numbers using a limited number of decimal places.

Suppose a calculator displays:

x = 1.333333

The exact value might be:

x = 4/3

If we substitute the rounded decimal into an equation, the result may be slightly different from the expected constant.

For example:

3(1.333333) = 3.999999

rather than exactly 4.

This small difference is caused by rounding.

Good practice

Keep exact values when possible.

If decimal approximations are required, retain sufficient precision during calculations and round only the final result.

A very small numerical difference may be acceptable, but a large discrepancy usually indicates an error.

Technology in Real-World Applications

Business and Economics

A business can use spreadsheets to solve systems involving product quantities, prices, revenue, and production costs.

Engineering

Engineers use computer software to solve systems describing forces, electrical circuits, and structural behaviour.

Science

Scientists use systems of equations to analyse experimental data, mixtures, and physical relationships.

Transport

Transport planners can use mathematical models to analyse vehicle movement, scheduling, and resource allocation.

Computer Graphics

Three-dimensional computer graphics use systems of equations and matrix calculations to represent and transform objects.

Common Mistakes

Entering equations incorrectly

A missing negative sign or coefficient can completely change the answer.

Using an inappropriate graphing window

An intersection may exist outside the visible region.

Assuming two lines do not intersect because they appear parallel

Lines with very similar slopes may intersect far away.

Treating approximate values as exact

Graphical and numerical solutions may be rounded.

Confusing the coordinates

The solution (3, 5) means x = 3 and y = 5, not the reverse.

Assuming every calculator error means no solution

Some errors indicate a singular matrix or unsupported calculation rather than an inconsistent system.

Using a two-dimensional graphing tool for three-dimensional equations

Three-variable systems generally require a three-dimensional graphing application or another solving method.

Accepting a technological output without checking it

Always substitute the proposed values into the original equations.

Ignoring real-world restrictions

Negative quantities or fractional counts may be mathematically possible but contextually invalid.

Key Terms

• Graphing calculator: A device or application that displays mathematical graphs.

• Graphing software: Computer software used to represent equations visually.

• Intersection point: A point shared by two or more graphs.

• Simultaneous equations: Equations solved together to determine common variable values.

• Numerical solution: A solution expressed using numerical values, often approximations.

• Exact solution: A solution expressed without rounding.

• Approximation: A value close to, but not necessarily equal to, an exact value.

• Graphing window: The range of coordinates displayed on a graph.

• Matrix: A rectangular arrangement of numbers.

• Coefficient matrix: A matrix containing the coefficients of a system.

• Inverse matrix: A matrix that reverses the effect of multiplication by an invertible matrix.

• Singular matrix: A square matrix that does not have an inverse.

• Computer algebra system: Software capable of performing symbolic mathematical operations.

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

• Consistent system: A system with at least one solution.

• Inconsistent system: A system with no solution.

• Dependent equations: Equations that contain redundant information.

• Rounding error: A small numerical difference caused by using approximate values.

Key Takeaways

• Technology can help solve and visualise systems of equations.

• Graphing software identifies solutions as intersection points.

• Two-variable linear systems are represented by lines.

• Three-variable linear systems are represented by planes.

• Scientific calculators can solve systems using coefficients and constants.

• Spreadsheets can use matrices to solve linear systems.

• Exact algebraic solutions and approximate numerical solutions should be distinguished.

• Graphing windows must be chosen carefully.

• Technology can reveal whether systems have one, no, or infinitely many solutions, although further analysis may be needed.

• Manual methods remain important for understanding and explaining mathematical relationships.

• Every technological solution should be verified using the original equations.

Check Your Understanding

Recall and Understanding

1. Explain two advantages of using technology to solve systems of equations.

2. What does the intersection of two lines represent?

3. Why might a graphing calculator fail to display an intersection even when one exists?

4. Explain the difference between an exact solution and an approximate solution.

5. Why is verification necessary when using technology?

Application

6. Use a graphing calculator or algebraic method to solve:

y = 2x + 3

y = −x + 12

7. Solve and verify:

3x + 2y = 16

x − y = 2

8. Determine the number of solutions:

y = 4x + 1

y = 4x − 3

9. Determine the number of solutions:

2x + 4y = 12

x + 2y = 6

10. Solve the three-variable system:

x + y + z = 12

2x + y + z = 15

x + 2y + z = 16

Analysis and Challenge

11. A graphing calculator displays the intersection (2.6667, 3.3333). Explain why these coordinates may not be exact.

12. A student enters the system:

2x + 3y = 12

x − y = 1

The calculator gives x = 3 and y = 2. Verify whether this is correct.

13. Explain why a system with dependent equations may cause a matrix-inverse calculation to fail.

14. A calculator produces x = −5, y = 20, z = 15 for a ticket-sales problem. Explain what should be checked before accepting the result.

15. Solve using a calculator, spreadsheet, or elimination:

2x + y + z = 9

x + 3y + 2z = 13

3x + 2y + z = 11

Answers to Check Your Understanding

1. Technology can perform complicated calculations quickly and display graphs that help us understand the relationships between equations.

2. The intersection represents the ordered pair that satisfies both equations simultaneously.

3. The intersection may lie outside the selected graphing window.

4. An exact solution is not rounded, such as 4/3. An approximate solution is a numerical estimate, such as 1.3333.

5. Equations may have been entered incorrectly, numerical results may be rounded, or the output may not make sense in the original context.

6.

Set the equations equal:

2x + 3 = −x + 12

3x = 9

x = 3

Then:

y = 2(3) + 3 = 9

Answer: (3, 9).

7.

3x + 2y = 16

x − y = 2

From the second equation:

x = y + 2

Substitute:

3(y + 2) + 2y = 16

5y + 6 = 16

y = 2

x = 4

Verification:

3(4) + 2(2) = 16

4 − 2 = 2

Answer: (4, 2).

8. No solution. The lines have the same slope but different y-intercepts.

9. Infinitely many solutions. The second equation multiplied by 2 gives the first equation, so both represent the same line.

10.

x + y + z = 12

2x + y + z = 15

x + 2y + z = 16

Subtract the first equation from the second:

x = 3

Subtract the first equation from the third:

y = 4

Substitute:

3 + 4 + z = 12

z = 5

Answer: (3, 4, 5).

11. The calculator may have rounded the coordinates. The exact values could be fractions such as 8/3 and 10/3.

12.

Check the first equation:

2(3) + 3(2) = 12

Correct.

Check the second:

3 − 2 = 1

Correct.

Answer: (3, 2) is a valid solution.

13. Dependent equations can produce a singular coefficient matrix. A singular matrix has no inverse, so the inverse-matrix method cannot be applied directly.

14. The values should be substituted into the original equations. Since x = −5, the result is invalid if x represents a number of tickets. Ticket quantities must be nonnegative whole numbers.

15.

The system is:

2x + y + z = 9

x + 3y + 2z = 13

3x + 2y + z = 11

From the first equation:

z = 9 − 2x − y

Substitute into the second:

x + 3y + 2(9 − 2x − y) = 13

−3x + y = −5

y = 3x − 5

Substitute into the third:

3x + 2y + 9 − 2x − y = 11

x + y = 2

Substitute y = 3x − 5:

4x − 5 = 2

x = 7/4

Then:

y = 3(7/4) − 5 = 1/4

And:

z = 9 − 2(7/4) − 1/4 = 21/4

Verification:

2(7/4) + 1/4 + 21/4 = 9

7/4 + 3(1/4) + 2(21/4) = 13

3(7/4) + 2(1/4) + 21/4 = 11

Answer: (7/4, 1/4, 21/4).

Final Summary

Graphing calculators, mathematical software, and spreadsheets provide powerful ways to solve systems of equations and investigate their properties.

Technology is particularly valuable for complex calculations, visualising intersections, and checking answers. However, understanding substitution, elimination, and the meaning of solutions remains essential.

The most effective mathematical problem-solving combines the efficiency of technology with the accuracy, interpretation, and reasoning provided by algebraic methods.

5. Error Analysis

Error Analysis: Solving Systems of Equations

Introduction

Solving systems of equations requires more than knowing how to use substitution, elimination, or graphing. It also requires the ability to recognise mistakes, explain why they occur, and correct them.

Even a small algebraic error can produce a completely different solution. A missing negative sign, an incorrectly distributed coefficient, or a substitution error may cause an otherwise correct method to fail.

For example, consider the system:

x + y = 10

x − y = 4

The correct solution is:

x = 7

y = 3

However, a student who makes a sign error during elimination might obtain x = 3 and y = 7.

Although these numbers add to 10, they do not satisfy the second equation:

3 − 7 = −4, not 4.

This demonstrates an important principle:

A solution to a system must satisfy every original equation, not just one of them.

Error analysis helps us become more accurate and independent mathematical problem-solvers. By examining incorrect work, we can understand why particular methods fail and develop strategies to prevent similar mistakes.

What Is Error Analysis?

Error analysis is the process of examining mathematical work to identify, explain, and correct mistakes.

It involves four important questions:

1. What was the student trying to do?

2. Where did the first mistake occur?

3. Why is that step incorrect?

4. How can the solution be corrected and verified?

The goal is not simply to identify a wrong answer. It is to understand the reasoning that produced the mistake.

Three Types of Errors

Some mistakes involve more than one category.

For example, subtracting a negative incorrectly may be an arithmetic mistake, but repeated errors could indicate a conceptual misunderstanding of negative numbers.

A Systematic Method for Finding Errors