# Linear Equation in One Variable Class 7

A linear equation is an algebraic equation consist variables has an exponent of one.
If we graph these equations in xy-plane results in a straight line. Linear equations are easy to deal with than non-linear equations. With the help of algebraic and graphing methods we can solve these equations. In this section will help class 7 students to solve linear equations with one variable step by step with the help of some solved examples. Inspite, of you being a brilliant student, linear equations solved examples for class 7 will help you aim for the marks in your examination. The content on linear equations has been carefully compiled and developed keeping into consideration the latest CBSE syllabus.

## Solving Linear Equation in One Variable

Use simple arithmetic operational to solve linear equations in one variable. These equations can be solved in one step or in two steps.
Keep below points in mind while solving such equations:
Follow BODMAS rule to simplify the equations.

## Examples

Let us see with the help of some examples, how to solve linear equations in one variable.
Example 1: Solve the following equation:

x + 5x + 4 = 3 (2x - 1)

Solution:
Step 1: Write down the equation given

x + 5x + 4 = 3 (2x - 1)

We see that there is existence of variables on both sides of the equal to. Same goes with the numbers as they are present on either side of the equal to.

Step 2: Using distributive property on right side of the equation.

x + 5x + 4 = 3 (2x) - 3(1)

x + 5x + 4 = 6x - 3

Step 3: We need to separate the terms containing variables on one side and the constants on the other side of the equal to. Combining the like terms shall give us

x + 5x - 6x = - 3 - 4

Step 4: Solving either side of the equation. So, it becomes

x - x  = - 3 - 4

or 0 = -7 (which is not true)

We see that when we subtract $x$ from $x$ it gives us $0$. As a result there are no terms in $x$ left in the equation to solve. Hence, no solution

As there is no solution obtained in the given linear equation, we can conclude that the equations are inconsistent in form.

Example 2: Solve 4(x + 3) - 10 = 8($\frac{3x}{2}$ + 2)

Solution:
Step 1: Using distributive property on both sides of the equation.
4(x) + 4(3) = 8($\frac{3x}{2}$) + 8(2)
4x + 12 = 12x + 16
Step 2: Subtract 12 x from both the sides
4x + 12 - 12x = 12x + 16 - 12x
-8x + 12 = 16
Step 3: Subtract 12 from each side
-8x + 12 - 12 = 16 - 12
-8x = 4
Step 4: To isolate variable x, divide each side by -8, we get
-8x/-8 = 4/-8
x = -4/8 = -1/2
The solution of given equation is : x = -1/2

## Practice Problems

Below are some practice problems to check your knowledge.
Problem 1: Solve 4(x + 3) - 4 = 8($\frac{x}{2}$ + 1)
Problem 2: Find the value of m, 12m - 3(m + 10) = 4(m - 2) + 6
Problem 3: Solve for x; 2x + 1 = 3(m + 9) - 3