Linear Equations: Comprehensive Guide

Introduction

Forms & Types

Solution Methods

Applications

Practice Quiz

What are Linear Equations?

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable raised to the first power.

Standard Form:

In one variable: ax + b = 0 where a ≠ 0

In two variables: ax + by + c = 0 where a and b are not both 0

Key Properties:

The highest exponent of any variable is 1
No products of variables (like xy)
No variables in denominators
No variables under radical signs

Why Linear Equations Matter

Linear equations form the foundation of algebra and are essential in:

Modeling direct relationships
Business applications (cost, revenue, profit)
Science and engineering applications
Data analysis and statistics
Computer graphics and animations

Forms and Types of Linear Equations

1. One-Variable Linear Equations

Standard Form: ax + b = 0 (where a ≠ 0)

Example: 3x + 7 = 16

Solution: 3x = 9 ⟹ x = 3

2. Two-Variable Linear Equations

Standard Form: ax + by + c = 0 (where a and b are not both 0)

Example: 2x + 3y - 6 = 0

3. Slope-Intercept Form

Form: y = mx + b

where m is the slope and b is the y-intercept

Example: y = 2x + 5

This line has slope 2 and y-intercept 5

4. Point-Slope Form

Form: y - y₁ = m(x - x₁)

where m is the slope and (x₁, y₁) is a point on the line

Example: y - 3 = 4(x - 2)

This line passes through point (2, 3) with slope 4

5. Intercept Form

Form: x/a + y/b = 1

where a is the x-intercept and b is the y-intercept

Example: x/4 + y/6 = 1

This line has x-intercept 4 and y-intercept 6

6. Systems of Linear Equations

A set of two or more linear equations considered simultaneously:

3x + 2y = 8

x - y = 1

7. Parametric Form

Form: x = x₀ + at, y = y₀ + bt

where (x₀, y₀) is a point on the line, (a, b) gives the direction, and t is a parameter

Example: x = 1 + 2t, y = 3 + 4t

Methods for Solving Linear Equations

1. One-Variable Linear Equations

Isolation Method (Most Common)

Solve for the variable by isolating it on one side of the equation.

Example: Solve 3x - 7 = 14

Add 7 to both sides: 3x = 21
Divide both sides by 3: x = 7

Balancing Method

Adding/subtracting the same number or multiplying/dividing both sides by the same non-zero number.

Example: Solve 2(x + 3) - 5 = x - 4

Expand: 2x + 6 - 5 = x - 4
Simplify: 2x + 1 = x - 4
Subtract x from both sides: x + 1 = -4
Subtract 1 from both sides: x = -5

2. Systems of Linear Equations

Substitution Method

Solve for one variable in terms of the other, then substitute.

Example: Solve the system:

x + y = 5

2x - y = 4

From first equation: y = 5 - x
Substitute into second: 2x - (5 - x) = 4
Simplify: 2x - 5 + x = 4
Solve: 3x = 9 ⟹ x = 3
Substitute back: y = 5 - 3 = 2
Solution: (3, 2)

Elimination Method

Add or subtract equations to eliminate a variable.

Example: Solve the system:

3x + 2y = 8

5x - 2y = 4

Add the equations (eliminates y):
3x + 2y = 8
5x - 2y = 4
8x = 12
x = 1.5
Substitute back: 3(1.5) + 2y = 8
4.5 + 2y = 8
2y = 3.5
y = 1.75
Solution: (1.5, 1.75)

Graphical Method

Graph both equations and find intersection point(s).

Example: Graphing y = 2x - 1 and y = -x + 4 shows they intersect at (1, 3)

Matrix Method (Cramer's Rule)

Using determinants to solve systems of equations.

For a system ax + by = e and cx + dy = f:

x = (e·d - b·f)/(a·d - b·c)

y = (a·f - e·c)/(a·d - b·c)

Applications of Linear Equations

1. Word Problems

Age Problems

Example: John is 5 years older than Mary. In 8 years, John will be twice as old as Mary was 2 years ago. How old are they now?

Let Mary's current age = x

John's current age = x + 5

In 8 years, John will be (x + 5 + 8) = x + 13

Mary 2 years ago was (x - 2)

Twice Mary's age 2 years ago = 2(x - 2) = 2x - 4

So: x + 13 = 2x - 4

Solve: 17 = x

Mary is 17, John is 22

Mixture Problems

Example: How many liters of a 20% acid solution must be mixed with 5 liters of a 60% acid solution to get a 30% acid solution?

Let x = liters of 20% solution

Amount of acid in 20% solution: 0.20x

Amount of acid in 60% solution: 0.60 × 5 = 3

Amount of acid in final 30% solution: 0.30(x + 5)

So: 0.20x + 3 = 0.30(x + 5)

Solve: 0.20x + 3 = 0.30x + 1.5

1.5 = 0.10x

x = 15

Need 15 liters of the 20% solution

2. Financial Applications

Break-Even Analysis

Example: A company produces widgets with fixed costs of $5000 and variable costs of $25 per widget. Each widget sells for $45. How many widgets must be sold to break even?

Let x = number of widgets

Total cost: C(x) = 5000 + 25x

Total revenue: R(x) = 45x

Break-even point occurs when C(x) = R(x)

5000 + 25x = 45x

5000 = 20x

x = 250 widgets