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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

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?

Solution:

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?

Solution:

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?

Solution:

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

Therefore,

x – 1/8x = 4

7/8x = 4

x = 32/7 tiles

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

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