RD Sharma Class 8 Solutions Chapter 9 - Linear Equation In One Variable (Ex 9.1) Exercise 9.1 - Free PDF

In this Chapter, you will be taught algebraic equations and expressions. We will be dealing with linear expressions in one variable only. Thus, such equations are known as linear equations in one variable.

These are of the format shown below:

Equations

• 5x + 2 = 7

• x =17

• 7y +1 = 8

We can clearly see that equations have an equality sign (=), which is not found in expressions from the illustrations. 

Variables, constants, and some mathematical operations like addition or multiplication are involved in an algebraic expression.

An expression that equates two expressions is an equation.

Some Points We Must Keep in Our Mind:

• An equality involving variables is called an algebraic equation. It will always have an equality sign. The part to the left of the equality sign is called the Left Hand Side or (LHS). The part to the right of the equality sign is called the Right Hand Side or (RHS).

• The values of the expressions on the LHS and RHS will always be equal in an equation. This is valid only for some values of the variable. These values are known as the solutions to the equation. 

• In order to find the solution to an equation, we assume two sides of the equation to be balanced. We then do the same operations on both sides of the equation to get the solution.

Some of the Topics which are discussed in this Chapter are as Follows

• Equations With Linear Expression on One side and Numbers on the Other Side

• Some Applications

• Equations that Have Variables on Both Sides

• Reducing Equations to Simpler Forms

Equations With Linear Expression on One side and Numbers on the Other Side

Let us learn this topic by solving equations like the one shown below:

5y - 12 = 8

To solve, will  by add 12 on both sides:

5y - 12 + 12 = 8 + 12

5y = 20

Hence,

y = 4

The above example is a linear expression with the highest power of a variable as only 1. There can be one or more than one variable in a linear equation.

Some Applications

There are applications of linear equations. They help us in the real world. There are many examples that involve some real-life situations like counting money, calculating age, finding perimeter and area etc.

Illustration: 

Question: Alexa is twice the age of Jenna. Ten years ago her age was three times

Jenna’s age. What are their current ages?

Solution: 

Let Jenna’s present age be x years.

Therefore Alexa’s present age will be 2x years.

Jenna’s age ten years ago was 

(x – 10) years.

Alexa’s age ten years ago was 

(2x – 10) years.

It is given that Alexa’s age ten years ago was three times Jenna’s age.

Thus, 

2x – 10 = 3(x – 10)

2x – 10 = 3x – 30

3x – 2x=30 – 10 

x=20

So, 

Jenna’s present age x = 20 years

While,

Alexa’s present age is 2x = 2 × 20 = 40 years.

Equations that Have Variables on Both Sides

So far, we have only seen equations where the values on the right-hand side of the equality sign have been numbers. Now, let us look into questions where there are variables on both sides.

Example:

3x - 7 = x +3

Solution: 

Adding 7 to both sides,

3x-7+7 =x + 3 + 7

Subtracting x from both sides we get 

3x - x = x + 10 - x

2x=10

Hence,

 x = 5

Reducing Equations to Simpler Forms

A complex linear equation with fractions can be reduced into simpler forms by the following steps:

• First, the LCM of the denominator is taken.

• Then, the RHS and LHS of the equation are multiplied both with the LCM.

• Therefore, the equation gets reduced to a form without a denominator in it.

To understand this better, let us take the help of an example:

\[\frac{x}{3}-\frac{1}{5}=\frac{x}{5}+\frac{1}{4}+2\]

\[\frac{5x}{15}-\frac{3x}{15}=\frac{4+5+40}{20}\]

\[\frac{2x}{15}=\frac{49}{20}\]

\[2x=\frac{49}{20}\times 15\]

\[x=\frac{49}{40}\times 15\]

\[x=\frac{147}{8}\]

Conclusion

For students who wish to score high marks in Math, RD Sharma Solutions is the best study material. The subject matter experts at Vedantu have prepared the  RD Sharma solutions to help the students who are finding difficulties in solving them. Students can easily access answers to the problems present in RD Sharma Class 8 Chapter 9  by downloading the PDF. It contains all solutions in a detailed manner and also expects questions to be asked in the exam. After solving these problems students will get more confident about the exam.