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Last updated on September 29, 2025
A linear equation is an equation made using variables, numbers, and operations having the highest power of 1. Solving a linear equation means finding the value of the variables that satisfy the equation.
A linear equation is an algebraic equation in which the highest power of the variable is 1. The values increase or decrease evenly, forming a straight-line graph. Coefficients multiply variables, while constants remain unchanged.
The solution to a system of linear equations can have three types of solutions, which are given below:
Unique Solution
A linear equation in one variable has a unique solution, where the LHS is equal to the RHS. A linear equation in two variables is represented on a graph by a straight line, and its solution is expressed as an ordered pair (x, y).
No Solution
In a linear equation, a system with no solution is also known as an inconsistent solution. The non-solution happens when two lines never intersect. In a system of two linear equations, if the slopes are the same, but the intercepts are different, the lines are parallel, and there will be no solution.
Infinite Solutions
The infinity solution occurs when the equations are on the same line. If both equations describe the same line, the lines overlap completely, and these are called coincident lines.
A linear equation contains only one variable in the expression. To solve a linear equation in one variable, keep the variable on one side and the constant on the other side. Then simplify the equation. For example,
3x - 7 = 5
Solution:
Keep the constant on one side and the variable on the other side,
3x -7 = 5
3x = 5 + 7
3x = 12
x = 12/3
x = 4
The value of x is 4.
How to Solve a Linear Equation in Two Variables?
The linear equation in two variables has two unknowns, usually x and y, and can be written as ax + by = c. For example, x + y = 6, find the value of x and y.
Solution:
We take the first two natural numbers instead of x
0:
x + y = 6
y = 6 -x
y = 6 -0
y = 6
x = 0
y = 6
1:
x + y = 6
y = 6 -1
y = 5
x = 1
y = 5
The methods used to solve linear equations are used in the solution of linear equations in two variables. There are several methods:
Substitution Method:
The substitution method is a way to solve linear equations by expressing one variable in terms of the other. Then substitute the expression into the second equation. This leaves an equation with only one variable, which helps to solve the equation and find the value.
For example,
x + y = 12 (1)
x = 4 (2)
Solution:
Substitute equation 2 in equation 1
x + y = 12
4 + y = 12
y = 12 -4
y = 8
So, x = 4 and y = 8.
Elimination Method
The elimination method is a method for solving linear equations. In this method, one of the variables is multiplied by a constant so that either the x term or the y term cancels out when the equations are added or subtracted, making it possible to find the value of the remaining variable.
For example,
x + y = 12 (1)
x - y = 4 (2)
Solution:
Add both equations
x + y = 12
x - y = 4
(x + y) + (x -y) = 12 + 4
2x = 16
y is eliminated because they both have a different sign
2x = 16
x = 16/2
x = 8
Substitute x = 8 in equation 1
x + y =12
8 + y = 12
y = 12 -8
y = 4
The values of x and y are 8 and 4
Graphical method
The graphical method is used to solve the linear equation, graphing the equations by finding the values of x and y in the coordinate system. After finding the value, find the intersection point of these two lines. The value of the point gives the solution for these linear equations. For example,
Find the intersection plot for lines:
y = x + 1
y = −x + 3
Solution
Slope intersection form y = mx + c
M is the slope
c is the intercept of y
Pick the values for x and calculate the y in equation 1
y = x + 1
-1 in x
y = -1 + 1 =0
y = 0
When x = 0,
y = 0 + 1
y = 1
1 in value of x
y = 1 + 1
y = 2
2 in value of x
y = 2 + 1
y = 3
(x, y) = (-1,0), (0,1), (1,2), (2,3)
Pick the values for x and calculate y in equation 2
y = −x + 3
When x = 0
y = 0 + 3 = 3
y = 3
1 in value of x
y = -1 + 3 = 2
y = 2
2 in value of x
y = -2 + 3 = 1
y = 1
3 in value of x
y = -3 + 3 = 0
y = 0
(x, y) = (0,3), (1,2), (2,1), (3,0)
Draw a straight line through the points.
Both lines intersect at the point:
(1,2)
Cross multiplication method
The cross multiplication method is used to solve linear equations by taking the coefficients of x, y, and the constant terms from the given equations.
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
The formation of the cross method is given below:
Let's see the example using the cross multiplication method
2x + 3y = 17
3x −2y = 6
Solution:
Write in the form like
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
2x + 3y - 17= 0 (a1 = 2, b1 = 3, c1 = -17)
3x −2y - 6 = 0 (a2 = 3, b2 = 2, c2 = -6)
x = (3 × -6 - 2 ×-17)
y = (-17 × 3 - (-6) ×2)
1 = (2 × 2 - 3 ×3)
x = -18 -34 = -52
y = -51 + 12 = -39
1 = -4 -9 = -13
Now compare
x = -52/-13 = 4
y = -39/-13 = 3
Linear equations are not only for academia, but it is also used in our day-to-day lives. Here are some real-life applications where linear equations are required.
Solving linear equations is an important part of algebra. While the steps to solve them may seem simple, students often make small mistakes, which lead to incorrect answers. These mistakes usually happen due to sign errors, skipping steps, or misunderstanding the rules of solving equations. Here are some common mistakes that are given below to help with a better understanding.
Solve: 3x −7 = 11
x = 6
Keep the constant on the other side
3x - 7 = 11
3x = 11 + 7
3x = 18
Divide both sides by 3:
3x = 18
x = 18/3
x = 6
Solve: 5x + 2 = 3x + 10
x = 4
Move the variables to one side and the constants to the other side
5x + 2 = 3x + 10
5x - 3x = 10 -2
2x = 8
Divide by 2
x = 8/2
x = 4
Solve: x + y = 7 (1), x - y = 1 (2)
x = 4, y = 3
(i) x + y = 7
(ii) x - y = 1
From (ii), we get
x = y + 1
Substitute this value of x in (i)
(y + 1) + y = 7
2y + 1 = 7
2y = 6
y = 6/2
y = 3
Now we substitute y = 3 in (ii)
x - 3 = 1
x = 3 + 1
x = 4
So, x = y and y = 3
Solve: 2(x−3) = x + 5
x = 11
Expand the brackets
2(x -3) = x + 5
2x -6 = x + 5
Move the variables to one side and the constants to the other side
2x -6 = x + 5
2x - x = 5 + 6
x = 11
Solve: x3 = 25
x = 1.2
Use the cross multiplication method:
x3 = 25
5x = 6
Divide by 5 on both sides:
5x = 6
x = 6/5
x = 1.2
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.