Chapter 3: Systems of Equations

Overview

Systems of equations are fundamental in Additional Mathematics, providing methods to solve problems with multiple unknown variables. This chapter explores both linear systems in three variables and simultaneous equations involving one linear and one non-linear equation. Mastery of these techniques is essential for solving complex problems in various fields including physics, engineering, and economics.

Learning Objectives

After completing this chapter, you will be able to:

Solve systems of linear equations in three variables
Apply elimination and substitution methods
Solve simultaneous equations with one linear and one non-linear equation
Interpret solutions geometrically
Apply systems of equations in real-world problem-solving

Key Concepts

3.1 Systems of Linear Equations in Three Variables

Definition

A system of three linear equations with three variables (x, y, z) has the general form:

\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}

The solution is a set of values (x, y, z) that satisfies all three equations simultaneously.

Geometric Interpretation

Geometrically, each equation represents a plane in 3D space. The solution represents the point of intersection of these three planes.

Possible Cases:

Unique solution: Three planes intersect at a single point
No solution: Planes are parallel or form a triangular prism (no common intersection)
Infinite solutions: All three planes intersect along a line or coincide (same plane)

3.2 Simultaneous Equations

Linear and Non-Linear Systems

A system consists of:

One linear equation: $ax + by = c$
One non-linear equation: such as quadratic ( $y = ax^2 + bx + c$ ), circle ( $x^2 + y^2 = r^2$ ), exponential, etc.

The solution represents the points of intersection between a line and a curve.

Important Formulas and Methods

Methods for Solving Systems of Linear Equations

1. Elimination Method

Steps:

Eliminate one variable to reduce to two equations
Solve the resulting 2×2 system
Substitute back to find the third variable

Example:

\begin{cases} x + y + z = 6 \quad (1) \\ 2x - y + 3z = 14 \quad (2) \\ x + 2y - z = 0 \quad (3) \end{cases}

Add (1) and (3): $2x + 3y = 6$ (4) Add (2) and (3): $3x + y = 14$ (5) Solve (4) and (5) to find x and y, then substitute into (1) to find z.

2. Substitution Method

Steps:

Express one variable in terms of others from one equation
Substitute into the other equations
Repeat until all variables are found

Example: From equation (1): $z = 6 - x - y$

Substitute into (2): $2x - y + 3(6 - x - y) = 14$ Simplify and solve for x and y.

Methods for Solving Simultaneous Equations

Substitution Method for Linear-Nonlinear Systems

Steps:

Express one variable from the linear equation
Substitute into the non-linear equation
Solve the resulting equation (usually quadratic)
Substitute back to find the other variable

Example: Given:

\begin{cases} x + y = 5 \quad \text{(linear)} \\ x^2 + y^2 = 13 \quad \text{(circle)} \end{cases}

From linear: $y = 5 - x$ Substitute: $x^2 + (5 - x)^2 = 13$ Solve quadratic: $2x^2 - 10x + 12 = 0$ Solutions: $x = 2, y = 3$ and $x = 3, y = 2$

Solved Examples

Example 1: System of Three Linear Equations

Solve the system:

\begin{cases} x + 2y - z = 4 \quad (1) \\ 2x - y + z = 1 \quad (2) \\ 3x + y + 2z = 9 \quad (3) \end{cases}

Solution using Elimination:

Step 1: Add (1) and (2) to eliminate z: $(x + 2y - z) + (2x - y + z) = 4 + 1$ $3x + y = 5$ (4)

Step 2: Multiply (2) by 2 and add to (3): 2(2x - y + z) + (3x + y + 2z) = 2(1) + 9 4x - 2y + 2z + 3x + y + 2z = 11 7x - y + 4z = 11 (5)

Wait, let's use a better approach:

Step 1: From (1) and (2): Add (1) and (2): 3x + y = 5 (4)

Step 2: From (2) and (3): Multiply (2) by 2: 4x - 2y + 2z = 2 (2a) Add to (3): (4x - 2y + 2z) + (3x + y + 2z) = 2 + 9 7x - y + 4z = 11 (5)

This is getting complicated. Let's use substitution:

From (1): z = x + 2y - 4 Substitute into (2): 2x - y + (x + 2y - 4) = 1 3x + y - 4 = 1 3x + y = 5 (4)

Substitute into (3): 3x + y + 2(x + 2y - 4) = 9 3x + y + 2x + 4y - 8 = 9 5x + 5y = 17 (5)

Now solve (4) and (5): From (4): y = 5 - 3x Substitute into (5): 5x + 5(5 - 3x) = 17 5x + 25 - 15x = 17 -10x = -8 x = 0.8

Then y = 5 - 3(0.8) = 5 - 2.4 = 2.6 Then z = 0.8 + 2(2.6) - 4 = 0.8 + 5.2 - 4 = 2

Solution: x = 0.8, y = 2.6, z = 2

Example 2: Simultaneous Linear and Quadratic Equations

Solve:

\begin{cases} y = 2x + 1 \quad \text{(linear)} \\ y = x^2 + 2x - 3 \quad \text{(quadratic)} \end{cases}

Solution:

Set equal since both equal y: $2x + 1 = x^2 + 2x - 3$ $0 = x^2 - 4$ $x^2 = 4$ $x = \pm 2$

For $x = 2$ : $y = 2(2) + 1 = 5$ For $x = -2$ : $y = 2(-2) + 1 = -3$

Solutions: $(2, 5)$ and $(-2, -3)$

Example 3: Linear and Circle Equations

Solve:

\begin{cases} x + y = 4 \quad \text{(linear)} \\ x^2 + y^2 = 8 \quad \text{(circle)} \end{cases}

Solution:

From linear: $y = 4 - x$ Substitute into circle: $x^2 + (4 - x)^2 = 8$ $x^2 + 16 - 8x + x^2 = 8$ $2x^2 - 8x + 16 = 8$ $2x^2 - 8x + 8 = 0$ $x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0$ $x = 2$

Then $y = 4 - 2 = 2$

Solution: $x = 2, y = 2$ (tangent point)

Example 4: Word Problem Application

A company produces two products A and B. The profit from A is RM20 per unit and from B is RM15 per unit. The production constraints are:

Total production time: 6x + 4y ≤ 24 hours
Material constraint: 2x + 3y ≤ 15 kg
Non-negativity: x ≥ 0, y ≥ 0

Find the maximum profit.

Solution:

First, find boundary points by solving equalities:

6x + 4y = 24 ⇒ 3x + 2y = 12
2x + 3y = 15

From equation 1: y = (12 - 3x)/2 Substitute into equation 2: 2x + 3[(12 - 3x)/2] = 15 2x + (36 - 9x)/2 = 15 Multiply by 2: 4x + 36 - 9x = 30 -5x = -6 x = 6/5 = 1.2 y = (12 - 3(1.2))/2 = (12 - 3.6)/2 = 8.4/2 = 4.2

Also check intercepts:

When x = 0: from 2x + 3y = 15, y = 5
When y = 0: from 6x + 4y = 24, x = 4

Corner points: (0,0), (4,0), (1.2,4.2), (0,5) Profit function: P = 20x + 15y

Calculate profit at each point:

(0,0): P = 0
(4,0): P = 80
(1.2,4.2): P = 20(1.2) + 15(4.2) = 24 + 63 = 87
(0,5): P = 75

Maximum profit is RM87 at (1.2,4.2)

Mathematical Derivations

Geometric Interpretation

In 3D space, each linear equation ax + by + cz = d represents a plane. The solution to a system of three equations is the intersection point of these three planes.

Possible Cases:

Unique solution: Three planes intersect at a single point
No solution: Planes are parallel or form a triangular prism
Infinite solutions: All three planes intersect along a line or coincide

Consistency of Systems

A system is:

Consistent: Has at least one solution
Inconsistent: Has no solution
Dependent: Has infinitely many solutions

Real-World Applications

1. Physics: Forces in Equilibrium

For a body in equilibrium:

ΣF_x = 0 (horizontal forces)
ΣF_y = 0 (vertical forces)
Στ = 0 (torques)

Example: A 10kg weight suspended by two cables at angles:

Cable 1: 30° to horizontal, tension T1
Cable 2: 45° to horizontal, tension T2

Equations: T1 cos 30° + T2 cos 45° = 0 (horizontal equilibrium) T1 sin 30° + T2 sin 45° = 98.1 (vertical equilibrium, mg = 10×9.81)

2. Chemistry: Chemical Reactions

For the reaction: aA + bB → cC + dD At equilibrium: a[A] + b[B] = c[C] + d[D]

Example: For $N_2$ + 3 $H_2$ → 2N $H_3$ at equilibrium: [ $N_2$ ] + 3[ $H_2$ ] = 2[N $H_3$ ] With initial concentrations, solve for equilibrium concentrations.

3. Economics: Supply and Demand

Market equilibrium occurs where supply = demand.

Example: Supply: P = 2Q + 10 Demand: P = 50 - 3Q Set equal: 2Q + 10 = 50 - 3Q 5Q = 40 ⇒ Q = 8, P = 26

Complex Problem-Solving Techniques

Problem: Find values of k for which the system has:

a) Unique solution b) No solution c) Infinite solutions

x + 2y = 4
2x + ky = 8

Solution:

For system ax + by = c and dx + ey = f:

Unique solution if ae - bd ≠ 0
No solution if ae - bd = 0 and af - cd ≠ 0
Infinite solutions if ae - bd = 0 and af - cd = 0

Here: a = 1, b = 2, c = 4, d = 2, e = k, f = 8

ae - bd = (1)(k) - (2)(2) = k - 4 af - cd = (1)(8) - (4)(2) = 8 - 8 = 0

a) Unique solution: k - 4 ≠ 0 ⇒ k ≠ 4

b) No solution: k - 4 = 0 and 0 ≠ 0 ⇒ Not possible

c) Infinite solutions: k - 4 = 0 and 0 = 0 ⇒ k = 4

Problem: Three planes intersect at a point. Find the point of intersection.

x + y + z = 3
2x - y + 3z = 7
x - 2y + z = 0

Solution:

Use elimination:

Add equations (1) and (2): (x + y + z) + (2x - y + 3z) = 3 + 7 3x + 4z = 10 (4)

Add equations (1) and (3): (x + y + z) + (x - 2y + z) = 3 + 0 2x - y + 2z = 3 (5)

Now we have: 3x + 4z = 10 (4) 2x - y + 2z = 3 (5)

From (1): y = 3 - x - z Substitute into (5): 2x - (3 - x - z) + 2z = 3 2x - 3 + x + z + 2z = 3 3x + 3z = 6 (6)

Now solve (4) and (6): From (6): x + z = 2 ⇒ x = 2 - z Substitute into (4): 3(2 - z) + 4z = 10 6 - 3z + 4z = 10 z = 4 Then x = 2 - 4 = -2 Then y = 3 - (-2) - 4 = 3 + 2 - 4 = 1

Solution: x = -2, y = 1, z = 4

Summary Points

Systems of equations solve problems with multiple unknowns
Three-variable systems can be solved by elimination or substitution
Geometric interpretation helps visualize solutions
Simultaneous equations find intersection points
Real-world applications include physics, chemistry, and economics
Consistency determines whether solutions exist

Common Mistakes to Avoid

Arithmetic errors - Carefully track signs and coefficients
Incomplete substitution - Substitute all occurrences of the variable
Geometric misinterpretation - Remember what intersection points represent
Word problem setup - Correctly translate words to equations
Checking solutions - Always verify solutions in all original equations

SPM Exam Tips

Exam Strategies

Master both methods - Elimination and substitution for different scenarios
Practice with word problems - Translate real situations to equations
Visualize geometrically - Understand what intersection means
Check consistency - Determine if solutions exist before solving
Work systematically - Organize your work to avoid confusion

Key Exam Topics

Three-variable linear systems (30% of questions)
Simultaneous linear-quadratic systems (40% of questions)
Word problems (20% of questions)
Geometric interpretation (10% of questions)

Time Management Tips

Basic three-variable systems: 6-8 minutes
Simultaneous equations: 5-7 minutes
Word problems: 8-10 minutes
Complex systems: 10-12 minutes

Practice Problems

Level 1: Basic Systems

Solve using elimination:

x + y + z = 6
2x - y + z = 3
x + 2y - z = 2

Solve using substitution:
```
x + y = 5
y + z = 7
z + x = 6
```

Level 2: Simultaneous Equations

Linear and quadratic:
```
y = x + 2
y = $x^2$ + 3x + 2
```
Linear and circle:
```
x + y = 6
$x^2$ + $y^2$ = 18
```
Linear and exponential:
```
y = 2x + 1
y = 3^(x-1)
```

Level 3: Applications

Investment Problem: RM10,000 invested at two interest rates: x% and y%. Total annual interest is RM850. If amount at x% is three times amount at y%, find x and y.
Mixture Problem: Two alloys containing gold and silver. Alloy A: 60% gold, 40% silver. Alloy B: 40% gold, 60% silver. How much of each to make 100kg of alloy that is 50% gold?

Physics Problem: Forces acting on a point:

$F_1$ cos 30° + $F_2$ cos 45° = 50
$F_1$ sin 30° + $F_2$ sin 45° = 100

Find $F_1$ and $F_2$ .

Did You Know? 📚

The method of elimination for solving systems of equations was developed by Chinese mathematicians as early as the 1st century CE. The famous "Nine Chapters on the Mathematical Art" contains systematic methods for solving linear systems using what is now called Gaussian elimination, predating Western mathematics by over 1500 years.

Quick Reference Guide

Concept	Method	Key Points
Three-variable system	Elimination/Substitution	Reduce to 2×2 then solve
Simultaneous equations	Substitution	Set equal or substitute one into another
Geometric interpretation	Planes/curves	Intersection points are solutions
Consistency	Determinant analysis	Check if solutions exist
Word problems	Translation	Convert words to equations

Systems of equations are powerful tools for solving real-world problems with multiple variables. Master these techniques as they form the foundation for linear algebra and optimization problems in higher mathematics.