We all use linear equations in our daily life without knowing it. Linear equations include untopicIdentified quantities or algebraic terms in the form of one or more variables and constant(s). In previous articles, we have already known what a linear equation is and what its types are. Now, we will find how to solve a linear equation.

Problems based on Addition and subtraction of a linear equation:

Let us start with a simple problem. 

Suppose the equation is x + 7= 22

In order to find the solution of X, first, we will subtract seven from both the stopicIde

x+ 7 -7 = 22 - 7

Result will be x = 15

Hence, the solution for the equation x + 7 = 22, the x = 15

Isn’t this simple!

Practice sums 

Example 1: Solve x-18= 18 

x= 18 + 18

x= 36

Example 2: Solve 10y = 9y- 3

10y - 9y= -3

y= -3

Example 3: Solve the equation a + 5 = -5 

a = -5 + 5 

a= 0

Problems based on multiplication and division of equation: 

Suppose the equation is (x/3) = 5

Here, the relation of 3 with x is in the form of division. If we take 3from the left-hand stopicIde of the equation to the right-hand stopicIde of the equation, it will be multiplied by 5.  

In other words, we have multiplied 3 on the left-hand stopicIde and right-hand stopicIde of the equation.

So we get, x= 5*3 = 15 

Hence, the solution for the equation (x / 3) = 5 is x = 15 

So, whenever the division with the variable is given, changing the stopicIde of the constant will make it get multiplied with a constant or algebraic term on RHS. 

Suppose the equation is 6x = 24

Here the relation of six with the variable x is in the form of multiplication. So, when we change the position of six to RHS, it will be divtopicIded by 24. In other words, we will divtopicIde 6 on both the right-hand stopicIde and the left-hand stopicIde of the equation.

So, x= 24/6 = 4 

Hence, the solution for the equation 6x = 24 is x = 4. 

So, whenever the multiplication with the variable is given, while changing its stopicIde it will become a division. 

Practice sums: 

Example 4: Solve   y / 7 = 2 / 21

y = 7 * 2 / 21 

y = 14/ 21 

y = 2 /3 

Example 5: Solve 14x = 2 

x = 2/14 

x= 1/7  

Mix bag: problems based on addition, subtraction, multiplication and division of equations: 

Example 6: Solve 5x/ 2 = 25/4

5x= 2 * 25 /4

5x= 50/4

x = 50 / 4*5

x= 10/4 

x= 5/2 

Example 7: Solve 4( x + 5) = 32

4x + 20 = 32

4x= 32 - 20 

4x= 12

x = 12/4 = 3

Example 8: Solve 5 + 2( a - 1) = 17

5 + 2a - 2 = 17

5 - 2 + 2a = 17

3 + 2a = 17

2a = 17 - 3

2a = 14 

a   = 14/ 2

a   = 7

Application in real life using equations: 

While using the equations in story sums or problem sums, we first need to understand the question and then we need to frame our equation. After framing the equation, we need to solve the equation. Now as we know problem sums of addition, subtraction, multiplication, and division of equations, we are ready to work the word problems of equations. Framing the equation is the main thing while solving problem sums. 

Example 9: Find a solution if four is added to six times a number and you get an answer 20. What will be that number? 

Let the number be x 

Six times x means: 6x 

Now, four is added to six times x and the result is 20. So we will frame this in the equation

6x + 4= 20 

As we are framing the equation now, we will solve it further.

6x = 20 - 4

x= 16/6 

x= 8/3 

So, the number is 8/3.