Students will learn the concepts more effectively with the help of the linear equation in one variable questions and answers. Almost every class includes a discussion of linear equations.. The questions will be prepared in accordance with the NCERT norms. Both in daily life and in mathematics, linear equations are used. So, it’s important to understand the fundamentals of linear equations in one variable. The problems presented here will cover both the fundamentals and more difficult problems for students of all skill levels. Click here to read more about linear equations in one variable.

What are Linear Equations in One Variable?

A linear equation in one variable is defined as an equation with only one variable. The standard form of linear equation in one variable is stated as Ax + B = 0, where A and B can be any two numbers, and x has just one possible value.

Also, read: Linear Equations.

Go through the below problems to understand the linear equation in one variable and also solve the practice problems.

Linear Equation in One Variable Questions with Solutions

1. Convert the given statement into equation: An integer increased by 5 equals 20.

Solution:

Given statement: An integer increased by 5 equals 20.

Assume that the unknown number is “x”.

Hence, x increased by 5 means x + 5.

Therefore, the equation x + 5 = 20

So, an integer increased by 5 equals 20 means x + 5 = 20, which is the required equation for the given statement.

2. Convert the given equation into a statement: 5x = 25.

Solution:

The equation given is 5x = 25.

The statement equivalent to the equation 5x = 25 is “Five times the number x equals 25”.

3. Find the value of x for the equation 2x – 7 = 21. Also, verify the answer.

Solution:

Given equation:2x – 7 = 21.

To solve the given equations, keep the variable x on one side and the constants on the other side.

Hence, the given equation becomes:

2x = 21 + 7

2x = 28

x = 28/2

x = 14

Therefore, the value of x is 14.

Verification:

Substitute x = 14 in the equation 2x – 7 = 21

Hence, 2(14) – 7 = 21

28 – 7 = 21

21 = 21

Hence, LHS = RHS.

4. Check whether x = 40 is the solution of the equation 5x/2 = 100.

Solution:

To check whether x = 40 is the solution or root of the equation 5x/2 = 100, put the value x = 40 in the equation 5x/2 = 100

⇒ [5(40)]/2 = 100

⇒ 200/2 = 100

⇒ 100 = 100

Since, LHS = RHS, x = 40 is the solution of the equation 5x/2 = 100.

5. Find the number, if 15 is added to three times of the number results in 45. Also, justify your answer.

Solution:

Given statement: 15 is added to three times of the number resulting in 45.

Let the unknown number be x.

Then according to the given statement, the equation formed is:

15 + 3x = 45

Now, keep the variable “x” on one side of the equation and constant on the other side of the equation.

⇒ 3x = 45 -15

⇒ 3x = 30

⇒ x = 30/3

⇒ x = 10

Therefore, the required number is x = 10.

Verification:

Substitute the value x =10 in the equation 15 + 3x = 45.

⇒ 15 + 3(10) = 45

⇒ 15 + 30 = 45

⇒ 45 = 45

Therefore, LHS = RHS.

Hence, verified.

Also, read: Linear Equations in Two Variables.

6. Check whether the given statements are true or false.

If x = 5, then 5x – 5 = 20
If x = 7, then 4x – 4 = 20.

Solution:

(a) Given equation: 5x – 5 = 20

If x = 5,

= 5(5) – 5

= 25 – 5

= 20

Hence, if x = 5, then 5x – 5 = 20.

Therefore, the given equation is true.

(b) Given equation: 4x – 4 = 20

If x = 7,

= 4(7) – 4

= 28 – 4

= 24

Hence, if x = 7, then 4x – 4 = 24.

Therefore, the given equation is false..

7. Determine the number, if the sum of two odd consecutive numbers is 56.

Solution:

As we know that the difference between two odd consecutive numbers is 2.

Hence, let the small odd numbers be x and the successive odd numbers is x+2.

According to the given condition, we can write

⇒ x + x + 2 = 56

⇒ 2x + 2 = 56

⇒ 2x = 56 – 2

⇒ 2x = 54

⇒ x = 54/2

⇒ x = 27, which is the required odd number.

Hence, the other odd number is:

⇒ x + 2 = 27 + 2 = 29.

Therefore, the two odd numbers are 27 and 29.

8. Solve the given linear equation: 17 + 6p = 9

Solution:

Given linear equation: 17 + 6p = 9

Now, keep the variable “p” on the left hand side and bring the constants on the right hand side.

Hence, we get

⇒ 6p = 9 – 17

⇒ 6p = -8

⇒ p = -8/6

⇒ p = -4/3

Therefore, the value of p is -4/3.

Also, read: How to Solve Linear Equations?

9. Find the two numbers, if the numbers are in the ratio 5 : 3 and they differ by 18.

Solution:

Let the unknown number be “p”.

Also, given that, the two numbers are in the ratio 5 : 3.

According to given conditions, we can write

⇒ 5p – 3p = 18

⇒ 2p = 18

⇒ p = 18/2

⇒ p = 9.

Therefore, the two numbers are:

5p = 5(9) = 45

3p = 3(9) = 27

Hence, the required two numbers are 45 and 27.

10. The ages of Rahul and Ramya are in the ratio 5: 7. After 4 years, the sum of their ages will be 56 years. Find their present ages.

Solution:

Assume that the ages of Rahul and Ramya are 5p and 7p.

After 4 years, the ages of Rahul and Ramya will be 5p + 4 and 7p + 4.

According to the given condition, we get the following equation:

(5p + 4) + (7p + 4) = 56

⇒ 12p + 8 = 56

⇒ 12p = 56 – 8

⇒ 12p = 48

⇒ p = 48/12

⇒ p = 4

Therefore, Rahul’s present age = 5(4) = 20

Ramya’s present age = 7(4) = 28.

Hence, the present age of Rahul and Ramya are 20 and 28, respectively.

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Practice Questions

Write the equation for the given statement: 3 decreased from 4x gives 17.
Convert the given equation into a statement: 9x – 9 = 81.
Determine the value of x for the given equation: 25x + 100 = 350.

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