Last modified on February 11th, 2025

chapter outline

 

Linear Equations

A linear equation is an algebraic equation in which the highest exponent of the variable is 1.

Here is an example of a linear equation:

7x + y = 8

This is the standard form of writing a linear equation. 

The graph of a linear equation always gives a straight line, as shown:

We can also write this equation in some other ways:

  • y = -7x + 8 (the Slope-Intercept Form)
  • y + 6 = -7(x – 2) (the Point-Slope Form)

However, these are NOT linear equations:

  • y = 5x2 – 2 (as the exponent of the variable x is 2)
  • x – y3 = 4 (as the exponent of the variable y is 3)
  • ${\sqrt{x}-y}$ = ${3}$ (as the exponent of the variable x is ${\dfrac{1}{2}}$)

Formulas

Linear equations can be grouped based on the number of variables.

In One Variable

The standard or general form of a linear equation in one variable is:

Ax + B = 0

Here,

  • A and B are the constants
  • x is the variable

In Two Variables

The standard form of a linear equation in two variables is:

Ax + By = C

Here

  • A, B, and C are the constants
  • x and y are the variables 

In Three Variables

The standard form of a linear equation in three variables is:

Ax + By + Cz = D

Here

  • A, B, C, and D are the constants
  • x, y, and z are the variables

System of Linear Equations

A system of linear equations contains two or more linear equations that involve the same set of variables. The solution to a system of linear equations is the set of values of the variables that satisfy all the equations simultaneously.

For example,

2x + 3y = 12 and x – y = 4 both are linear equations. Together, they form a system of linear equations. 

Solving

For One-Variable

When solving equations involving one variable, we first isolate the variable to one side and then solve for the variable. 

Let us solve 2x + 5 = 11

Subtracting 5 from both sides, we get

⇒ 2x + 5 – 5 = 11 – 5

⇒ 2x = 6

On dividing both sides by 2, we get

⇒ ${\dfrac{2x}{2}=\dfrac{6}{2}}$

⇒ x = 3

Thus, the variable x = 3

Solve the linear equation: 2x + 7 = 5x – 3

Solution:

Given, 2x + 7 = 5x – 3
⇒ 2x + 7 – 7 = 5x – 3 – 7
⇒ 2x = 5x – 10
⇒ 2x – 5x = 5x – 10 – 5x
⇒ -3x = -10
⇒ 3x = 10
⇒ ${\dfrac{3x}{3}=\dfrac{10}{3}}$
⇒ ${x=\dfrac{10}{3}}$
Thus, x = ${\dfrac{10}{3}}$

For Two-Variables

The two-variable linear equations can be solved either by substitution or elimination method:

  1. Substitution Method
  2. Elimination Method

Problem: Solving linear equations by SUBSTITUTION method

Solve: 
2x + y = 8 
x – y = 2

Solution:

Given,
2x + y = 8 …..(i)
x – y = 2 …..(ii)
First, we solve one equation for one variable and then use that solution to find the other variable.
From equation (ii), solving for x in terms of y, we get
x = y + 2
Substituting x = y + 2 into equation (i), we get
2(y + 2) + y = 8
⇒ 2y + 4 + y = 8
Now, simplifying, we get
⇒ 3y + 4 = 8
⇒ 3y = 8 – 4 
⇒ 3y = 4
⇒ y = ${\dfrac{4}{3}}$
Substituting y = ${\dfrac{4}{3}}$ back into x = y + 2, we get 
x = ${\dfrac{4}{3}+2}$
⇒ x = ${\dfrac{10}{3}}$
Thus, x = ${\dfrac{10}{3}}$ and y = ${\dfrac{4}{3}}$

Problem: Solving linear equations by ELIMINATION method

Solve: 
2x + y = 8 
x – y = 2

Solution:

Given,
2x + y = 8 …..(i)
x – y = 2 …..(ii)
In this method, we first eliminate one variable by adding or subtracting the equations and then solve for the remaining variable.
Aligning both equations, we get
2x + y = 8 …..(i)
x – y = 2 …..(ii)
Now, adding the equations to eliminate y, we get
(2x + y) + (x – y) = 8 + 2
⇒ 2x + y + x – y = 10
Simplifying further,
⇒ 2x + x = 10
⇒ 3x = 10
⇒ x = ${\dfrac{10}{3}}$
Now, substituting x = ${\dfrac{10}{3}}$ into one of the original equations, we get
x – y = 2
⇒ y = x – 2
⇒ y = ${\dfrac{10}{3}-2}$
⇒ y = ${\dfrac{4}{3}}$
Thus, x = ${\dfrac{10}{3}}$ and y = ${\dfrac{4}{3}}$

Solve the following system of linear equations by the elimination method:
3x + 2y = 12
5x – 2y = 8

Solution:

Given, 
3x + 2y = 12 …..(i)
5x – 2y = 8 …..(ii)
Adding equations (i) and (ii), we get
(3x + 2y) + (5x – 2y) = 12 + 8
⇒ 3x + 2y + 5x – 2y = 20
⇒ 3x + 5x = 20
⇒ 8x = 20
⇒ ${\dfrac{8x}{8}=\dfrac{20}{8}}$
⇒ ${x=\dfrac{5}{2}}$
Now, from equation (ii),
5x – 2y = 8
⇒ -2y = 8 – 5x
⇒ 2y = 5x – 8
⇒ 2y = ${5\left( \dfrac{5}{2}\right) -8}$
⇒ 2y = ${\dfrac{25}{2}-8}$
⇒ 2y = ${\dfrac{25-16}{2}}$
⇒ 2y = ${\dfrac{9}{2}}$
⇒ ${\dfrac{2y}{2}=\dfrac{9}{4}}$
⇒ ${y=\dfrac{9}{4}}$
Thus, x = ${\dfrac{5}{2}}$ and y = ${\dfrac{9}{4}}$

Note: Linear equations in two variables can also be solved graphically, by cross-multiplication, or by the determinant method.

Graphing

The graph of a linear equation with one variable, x, forms a vertical line parallel to the y-axis.

Similarly, the graph of a linear equation with one variable, y, forms a horizontal line parallel to the x-axis.

The graph of a linear equation of two variables, x and y, always gives a straight line.

This is because a linear equation represents a constant rate of change between the two variables. 

  • A horizontal line has the equation y = c, with m = 0
  • A vertical line has the equation x = a with an undefined slope

Let us graph the linear equation y = 2x + 3

Step 1: Rewriting in Slope-Intercept Form (If Necessary)

As we know, the slope-intercept form is y = mx + c

The given equation, y = 2x + 3, is already in slope-intercept form.

Here,

  • m = 2 
  • c = 3 

Step 2: Identifying the Slope and Y-Intercept

Y-Intercept (c): This is the point where the line crosses the y-axis. In this case, c = 3. Thus, the y-intercept is (0, 3)

Slope (m): The slope m = 2 means that if x increases 1 unit, y increases by 2 units (rise = 2, run = 1)

Step 3: Using the Slope to Find Another Point

Using the y-intercept and the slope, here are some more values:

x2x + 3y
-22(-2) + 3-1
-12(-1) + 31
02(0) + 33
12(1) + 35
22(2) + 37

Step 4: Plotting the Points on the Coordinate Plane

Special Cases: Graphing Identity and Constant Functions

Apart from graphing standard linear equations, we can also graph two special types of linear functions:

Identity Function (y = x) is a straight line passing through the origin (0, 0) at a 45° angle, with a slope of 1. Every input value equals the output value.

Constant Function (y = c) is a horizontal line parallel at y = c since the slope is 0. The value of y remains the same for all values of x. Here is a graph of y = 5 shown.

Graph the equation ${y=-\dfrac{2}{3}x+4}$

Solution:

Given, ${y=-\dfrac{2}{3}x+4}$
The slope is m = ${-\dfrac{2}{3}}$
The y-intercept is c = 4
Thus, the coordinates of the y-intercept are (0, 4)
Now, the other points are:
If x = -3, then y = 6
If x = 3, then y = 2
If x = 6, then y = 0
If x = 9, then y = -2
Thus, by plotting these points, we get the required graph.

A company rents cars for \$50 per day plus \$0.25 per mile driven. Write the linear equation to represent the total cost (c) for renting a car for one day and driving x miles. Then, find the cost if 120 miles are driven.

Solution:

As we know, the total cost is the sum of the daily charge and the cost per mile.
Thus, the linear equation is: c = 50 + 0.25x
If x = 120, then c = 50 + (0.25)(120)
⇒ c = 50 + 30
⇒ c = 80
Thus, the total cost is $80

Verify whether the point (2, 7) lies on the line y = 3x + 1.

Solution:

Given, y = 3x + 1
Here, x = 2 and y = 7
Now, 7 = (3)(2) + 1 = 6 + 1 = 7, the point (2, 7) satisfies the equation.
Thus, the point (2, 7) lies on the line y = 3x + 1

Last modified on February 11th, 2025