NCERT Solutions Class 7 Maths Chapter 4

NCERT Solutions for Class 7 Mathematics Chapter 4 

NCERT Solutions for Class 7 Mathematics Chapter 4 Simple Equations 

The NCERT Solutions for Class 7 Mathematics Chapter 4 by Extramarks is a compilation of detailed step-by-step solutions to all exercises included in this chapter.

Students should practise all the exercise questions to get in-depth knowledge about the topics. The solutions are crafted by the subject matter experts, who have framed solutions in a systematic and organised manner  which is easy to understand. Students who refer to these materials will be able to prepare confidently  for their exams and achieve desired  results. 

NCERT Solutions for Class 7 Mathematics Chapter 4 Simple equations

Access NCERT Solutions for Class 7 Mathematics Chapter 4 – Simple Equations 

NCERT Solutions for Class 7 Mathematics Chapter 4 

Chapter 4 of Class 7 Mathematics is on simple equations divided into five major sections. It is one of the most important chapters in Class 7 Mathematics, as it brushes-up  basic concepts of algebraic equations. 

Students are advised to go through the chapter to get a clear understanding of the concepts in simple equations. The solutions to the questions in this chapter provided by Extramarks will help students to clarify the concepts and they will be able to  solve any questions in the term tests and exams confidently. 

 Following are the important topics covered under NCERT Class 7 Mathematics Chapter 4.

1. Stepping up of an equation 
2. Review of what we know
3. What Equation is?
4. Solving an equation
5. More Equations
6.  From Solution to Equation
7. Application of Simple Equations to practical situations

NCERT Solutions for Class 7 Mathematics Chapter 4 Exercises 

The total number of questions in each of the chapter’s exercises are given in the table below.

Facts 

• A variable takes on different numerical values whereas a constant has a fixed value. 
• An equation is a statement of a variable in which two expressions of the variable should have equal value.
• An equation remains unchanged if its LHS and RHS are interchanged.
• Transposing a number means moving it to the other side.
• The equations remain unchanged when we:

• Add the same number to both sides. 
• Subtract the same number from both sides.
• Multiply  and divide both sides by the same number.

• When we transpose a number from one side of the equation to the other its sign changes

Variable 

A variable does not have a fixed value. The numerical value of the variable changes. These variables are denoted by letters of the alphabet such as l, m, n, p, q, r, s, t, u, v, w, x, y, z, etc. Expressions are formed when we perform operations such as addition, subtraction, multiplication, and division on variables. 

• The value of an expression depends upon the chosen value of the variable. If there is only one term in an expression then it is called a monomial expression.
• If there are two terms in an expression then it is called a binomial expression. 
• If there are three terms in an expression then it is called a trinomial expression. 
• A polynomial expression is an expression that has four terms.

Note: A polynomial expression can have many terms but none of the terms can have a negative exponent for any variable.

An Equation

An equation is a mathematical statement on a variable where two expressions on either side of the equal sign should have equal value. At least one of the expressions must contain the variable. 

Note: An equation does not change when the expression on the left-hand side or the right-hand side is interchanged. 

In an equation, there is always an equality sign between two expressions.

Example: Write the following statements in the form of equations.

1. The difference of five times x and 11 is 28.
2. One-fourth of a number minus 8 is 18.

Solution:

1. We have five times x that is 5x

The difference of 5x-11 is 5x-11

5x-11=28

Thus, the required equation is 5x-11=28

1. Let the number be x

One-fourth of x is ¼(x)

Now, one-fourth of x minus 8 is 1/4(x) – 8

Thus, the required equation is ¼(x) – 8=18

Let us see one more example which will help you with Exercise 4.1 of NCERT solutions chapter 4 

Example: Write a statement for the equation 2x-5=15

Solution: 2x-5=15

Taking away 5 from twice a number is 15

Solving an Equation 

We use this principle when we solve an equation. The equality sign between the LHS and RHS corresponds to the horizontal beam of the balance. 

An equation remains undisturbed or unchanged:

1. If LHS and RHS are interchanged.
2. To both the sides, if the same number is added
3. From both sides if the same number is subtracted.
4. When both LHS and RHS are multiplied by the same number 
5. When both LHS and RHS are divided by the same number

To understand the concept better, let us try to solve an example. This will help you with exercise 4.2 of NCERT Solutions Chapter 4.

Example: Solve 5x-3=12

Adding 3 to both sides, we get 

5x-3+3=12+3

5x=15

Dividing both sides by 5, we get 5x/5=15/5

x=3, which is the required solution.

Note: For checking  the answer, we substitute the value of the variable in the given equation

i.e., L.H.S = (5*3)-3= 15-3= 12= R.H.S

or L.H.S = R.H.S

Example: ½(x) + 5= 65

Subtracting 5 from both sides we have,

½(x) +5-5 = 65-5

½(x) = 60

Multiplying 2 on both sides, we have

½(x) *2 = 60*2

x = 120, is the required solution.

Forming an Equation

We have learned how to solve an equation. Now we shall form or construct the equation when the solution(root) is given. Let us know the following successive steps:

• Start with x = 9
• Multiply both sides by 3 

3x = 27

• Subtract 2 from both sides 

3x – 2 = 27-2

3x – 2 = 25, which is an equation.

Note: For a given equation, you get one solution; but for a given solution, one can make many equations. 

Let us understand this with  more examples so that you can solve exercise 4.3 of NCERT Solutions Chapter 4.

Example: Solve 5(x-3) = 25

(Or) x-3 = 25/5 (Dividing both sides by 5)

(Or) x – 3 = 5

(or) x = 5+3  (Transposing -3 to R.H.S)

x = 8, which is the required solution.

Example: Solve 3(x+1)/2 = 18

Solution: 3(x+1)/2 = 18

(or) (x+1)/2 = 18/2 (Dividing both sides by 2)

(or) (x+1)/2 = 6

(or) x/2 = (6-1)/2 (Transposing 1 to R.H.S)

(or) x = (12-1)/2 = 11/2, which is the required solution.

Application of Simple Equations to Practical Situations

Let us understand this with more examples so that you can solve exercise 4.4 of NCERT Solutions Chapter 4. 

Example: The sum of five times a number and 18 is 63. Find the number 

Let the required number be x 

5 times the number is 5x

According to the condition, we have 

5x + 18 = 63

5x = 63 – 18    (Transposing 18 form L.H.S to R.H.S)

5x = 45

(or) dividing both sides by 5, we have 

5x/5  = 45/5

x = 9

Thus, the required number is = 9

Related Questions 

Question: If 2x-3 = 5, then 

• X = 1
• X = -1
• X = 4
• X = -4

Solution: x = 4

Question: If both sides of the equation are divided by the same (non–zero) quantity, the equality –

• Does not change 

• Changes 

• May or may not change 

• None of these 

Solutions: Does not change

NCERT Solutions Class 7 Maths Chapter-wise List