Table of Contents
Last modified on February 11th, 2025
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:
However, these are NOT linear equations:
Linear equations can be grouped based on the number of variables.
The standard or general form of a linear equation in one variable is:
Ax + B = 0
Here,
The standard form of a linear equation in two variables is:
Ax + By = C
Here,
The standard form of a linear equation in three variables is:
Ax + By + Cz = D
Here,
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.
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
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 = ${\dfrac{10}{3}}$
The two-variable linear equations can be solved either by substitution or elimination method:
Problem: Solving linear equations by SUBSTITUTION method
Solve:
2x + y = 8
x – y = 2
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
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
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.
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.
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,
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:
x | 2x + 3 | y |
---|---|---|
-2 | 2(-2) + 3 | -1 |
-1 | 2(-1) + 3 | 1 |
0 | 2(0) + 3 | 3 |
1 | 2(1) + 3 | 5 |
2 | 2(2) + 3 | 7 |
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 ${y=-\dfrac{2}{3}x+4}$
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.
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.
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