MathIsimple
Lesson 1.4: Systems of Linear Equations

Systems of Linear Equations

Master two powerful methods for solving systems of linear equations: substitution and elimination, through engaging real-world scenarios.

What are Systems of Linear Equations?

Understanding the concept and types of solutions

Definition

A system of linear equations is a collection of two or more linear equations with the same variables that we solve simultaneously.

Example:
x + y = 5
2x - y = 1

Goal: Find values of x and y that satisfy BOTH equations

The solution is the point where the two lines intersect

Types of Solutions

One Solution: Lines intersect at one point

Example: x = 2, y = 3 (unique solution)

No Solution: Lines are parallel (never intersect)

Example: Inconsistent system

Infinitely Many Solutions: Lines are identical

Example: Dependent system

Method 1: Substitution

Solve one equation for one variable, then substitute into the other equation

Substitution Steps

1Solve for one variable

Choose the easier equation and solve for x or y

2Substitute

Replace the variable in the second equation

3Solve

Solve the resulting equation for the remaining variable

4Back-substitute

Use the found value to solve for the other variable

Example: x + y = 5, 2x - y = 1

Step 1: Solve first equation for y

x + y = 5 → y = 5 - x

Step 2: Substitute into second equation

2x - (5 - x) = 1

Step 3: Solve for x

2x - 5 + x = 1 → 3x = 6 → x = 2

Step 4: Find y

y = 5 - 2 = 3

Solution: (2, 3)

Check: 2 + 3 = 5 ✓, 2(2) - 3 = 1 ✓

Method 2: Elimination

Add or subtract equations to eliminate one variable

Elimination Steps

1Align equations

Write equations in standard form: ax + by = c

2Make coefficients equal

Multiply equations to make one variable's coefficients equal

3Add or subtract

Add or subtract equations to eliminate one variable

4Solve and substitute

Solve for one variable, then substitute to find the other

Example: 3x + 2y = 7, x - 2y = -1

Step 1: Equations are already aligned

3x + 2y = 7
x - 2y = -1

Step 2: y coefficients are opposites (2 and -2)

No multiplication needed!

Step 3: Add equations

(3x + 2y) + (x - 2y) = 7 + (-1)
4x = 6 → x = 1.5

Step 4: Substitute to find y

1.5 - 2y = -1 → -2y = -2.5 → y = 1.25

Solution: (1.5, 1.25)

Check: 3(1.5) + 2(1.25) = 7 ✓, 1.5 - 2(1.25) = -1 ✓

Real-world Application: Concert Tickets

Let's solve a practical problem using systems of equations

The Problem

Scenario: A concert venue sells two types of tickets:

  • • VIP tickets: $50 each
  • • General admission: $25 each

Given:

  • • Total tickets sold: 200
  • • Total revenue: $7,500

Find: How many of each type of ticket were sold?

Solution Using Substitution

Step 1: Define variables and write equations

Let x = VIP tickets, y = General tickets
x + y = 200 (total tickets)
50x + 25y = 7500 (total revenue)

Step 2: Solve first equation for y

y = 200 - x

Step 3: Substitute and solve

50x + 25(200 - x) = 7500
50x + 5000 - 25x = 7500
25x = 2500 → x = 100

Step 4: Find y

y = 200 - 100 = 100

Answer:

100 VIP tickets and 100 General tickets
Check: 100 + 100 = 200 ✓, 50(100) + 25(100) = 7500 ✓

Advanced System Concepts

Explore more complex aspects of systems of equations

Types of Solutions

One Solution: Lines intersect at one point

Example: x + y = 5, 2x - y = 1 → (2, 3)

No Solution: Parallel lines (never intersect)

Example: x + y = 3, x + y = 5 → Inconsistent system

Infinite Solutions: Same line (overlapping)

Example: x + y = 3, 2x + 2y = 6 → Dependent system

Matrix Method Preview

Augmented Matrix:

[1 1 | 5]
[2 -1 | 1]

Row Operations:

• Swap rows
• Multiply row by constant
• Add multiples of rows

Goal: Reduce to row-echelon form

[1 0 | 2]
[0 1 | 3] → x = 2, y = 3

Systems in Three Variables

Introduction to solving systems with three unknowns

Elimination Method

Step 1: Eliminate one variable

Use two equations to eliminate x, y, or z

Step 2: Create 2×2 system

Solve the resulting two-variable system

Step 3: Back substitute

Find the third variable using original equations

Example System

System:

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

Solution:

x = 1, y = 2, z = 3

Real Applications

Chemistry: Balancing equations

Find coefficients for chemical reactions

Economics: Supply and demand

Market equilibrium with multiple goods

Engineering: Circuit analysis

Current flow in electrical networks

Practice Problems

Apply both methods to solve these systems

Problem 1: Use Substitution

System:
y = 2x + 1
3x + y = 10

Solution:

Substitute: 3x + (2x + 1) = 10

5x + 1 = 10 → 5x = 9 → x = 1.8

y = 2(1.8) + 1 = 4.6

Answer: (1.8, 4.6)

Problem 2: Use Elimination

System:
2x + 3y = 13
2x - y = 1

Solution:

Subtract: (2x + 3y) - (2x - y) = 13 - 1

4y = 12 → y = 3

2x - 3 = 1 → 2x = 4 → x = 2

Answer: (2, 3)

Ready for the Final Lesson?

Now that you can solve systems of equations, let's apply linear functions to real-world scenarios.