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Linear Equations in One Variable – Solving Equations which have Linear Expressions on one Side and Numbers on the other Side | Class 8 Maths

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Linear equation is an algebraic equation that is a representation of the straight line. Linear equations are composed of variables and constants. These equations are of first-order, that is, the highest power of any of the involved variables i.e. 1. It can also be considered as a polynomial of degree 1. Linear equations containing only one variable are called homogeneous equations. The corresponding variable is called the homogeneous variable.

Solving a system of linear equations in one variable

The linear equations in one variable are represented in the form ax+b = 0 where x is a variable, a is a coefficient and b is a constant. These equations can be solved by the following steps: 

Step 1: In case the integers a and b are fractional numbers, LCM has to be taken to clear them.

Step 2: The constants are taken to the right side of the equation.

Step 3: All the terms involving the variable are isolated to the left-hand side of the equation, to evaluate the value of the variable.

Step 4: The solution is verified. 

Sample Problems on Linear Equations

The following examples illustrate the complete procedure of solving algebraic expressions composed of variables on one side (L.H.S) and constants on other (R.H.S).

Examples 1: For equation 5x – 20 = 100, find the value of x?


Transposing 20 to RHS, we have,
5x = 120

Dividing both side by 5,
5/5 x = 120/5
x = 24, which is the final solution

Example 2: For equation 4/3z + 1/9 = -1, find the value of z?


Transposing 1/9 to RHS,
4/3 z = -1 -1/9
⇒ 4/3 z = -8/9

Multiplying both sides by 3/4,
⇒ 3/4 x 4/3 z = -8/9 x 3/4
z = -2/3, which is final solution

Example 3: For equation 9/10 + y = 3/2, find the value of y?


Transposing 9/10 to RHS, we get, 
y = 3/2 – 9/10

Taking LCM of RHS
y = (15-9)/10
y = 6/10

Simplifying, we have
y = 3/5 , which is the final solution.

Note: The solution for the variable may be integers or rational numbers.

Example 4: Ben’s age is 4 times the age of William. 10 years earlier, Ben’s age was 14. What is his William’s current age?

Solution: Let William’s current age be w. 

Now , Ben’s current age is 4 times the age of William, which is equivalent to 4w. 

Ben’s age 10 years earlier = Ben’s current age – 10 

                                             = 4w -10 

Now, according to the question,

4w – 10 = 14

4w = 24

w = 6 years

Therefore, William’s current age is 6 years.

Example 5: Aman bought 5 chocolates for Rs 35. What is the cost of each chocolate?

Solution: Let the cost of each chocolate be r Rs. 

Now, according to the question,

5x = 35

x= Rs 7

Therefore, the cost of each chocolate is Rs 7. 

Example 6: Sita had some tiles and Gita had 1/8 of the tiles of Sita. The difference in their tiles is 4. What is the number of tiles with Gita?

Solution: Let the number of tiles with Sita be x.

Now, Gita has 1/8 x tiles.

According to the question , 

SIta’s tiles – Gita’s tiles = 4


 x – 1/8x = 4

7/8x = 4

x = 32/7 tiles

Now, Gita has 1/8 * 32/7 tiles = 4/7 tiles.

Last Updated : 31 Oct, 2020
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