# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (Ex 4.2) Exercise 4.2

On the Extramarks website and the mobile application, students can find the easy PDF solutions to the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations. In this resource provided by Extramarks, which is the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2, only one exercise, which is solving an Equation has been addressed. Students will learn how to work through the equations until they arrive at a solution. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations, have been created by knowledgeable teachers at Extramarks, and they are accurate, thorough, and informative.

The students should keep in mind that an equation is similar to a weighing scale when working with them. An operation on an equation is analogous to subtracting or adding weights on a scale. A Mathematical equation is analogous to a balance scale with equal weights on both of its plates. This is the main  idea taught in  NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2, which will aid the students in understanding how to solve an equation.

The practise questions in Extramarks’ NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2,  are mostly concerned with the processes that students will follow to solve the algebraic entities. In the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2, there are four problems that each comprise a number of short questions.

The equality of the equation will not hold if students do not carry out the exact same Mathematical procedure on both sides of an equilibrium. On the Extramarks website and mobile application, students can access the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 in PDF format.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (Ex 4.2) Exercise 4.2

Before moving on to the exercise questions, students must read through the examples in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations. In addition to the examples, real-world examples have helped to make the rationale behind the equation very evident in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2. Therefore, a thorough reading of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 will guarantee that the students have the requisite foundational knowledge and specifics on the solution of equations.

Students’ ability to solve the activities correctly depends on their number sense and mental arithmetic skills. To complete the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2, all that is needed is a firm grasp of the fundamental concepts of arithmetic.

For the benefit of the students, NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 is included as a PDF reference in the article and may be accessed both online and offline on the Extramarks website and mobile application. The principle of solving an equation is completely explained in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2. Students gain the ability to fully solve an equation and determine the solution.

Here is a summary of some key ideas to keep in mind when solving equations in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2:

• On both sides of the equation, adding the same number is possible (LHS and RHS).
• Both sides of the problem can be divided by the same number (LHS and RHS).
• A non-zero number can be multiplied or divided by itself on both sides of an equation (LHS and RHS).
• If the same operation is not performed on both sides, the equality of the equation is violated (LHS and RHS).

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

A Mathematical equation is a statement about a variable that specifies that two expressions on each side of the equality sign should have the same value. Two expressions along both sides of the sign are related in the straightforward equation. It has a single variable. The left-hand side and right-hand side are the names of the expressions on both sides (right-hand side). A constant’s value is fixed, whereas a variable’s value can change. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 will teach students what an equation is. Any equation has an equality sign that indicates the value of the expression to the left of the equation’s left sign is equal to the equation’s right side. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 will teach students how to compute or solve an equation.

Moreover, these NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 on Simple Equations address the idea that Algebra is the study of the unknown and that any condition on this unknown variable is referred to as an equation. Equations are described as situations where a variable must meet certain criteria and the values of the two expressions must match. The answer to the equation is revealed once the value of the variable that applies to it has been determined. The aforementioned facts are broadly discussed in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 offered by Extramarks.

### NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2  simple equations begin with the fundamentals of defining a variable and then show how an algebraic expression can be constructed around it and answered.

Students will learn that even if the right-hand side expression and the left-hand side expression of an equation are switched around, the equation will still hold true for any given set of circumstances. The link on the Extramarks website and mobile application will take students to the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations.

Simple Equations is the topic of Chapter 4 of the Class 7 Mathematics curriculum. Since it introduces and develops the idea of algebraic equations, this chapter on simple equations is an important part of the Class 7 Mathematics curriculum. This extremely important Mathematics chapter, which is addressed in Class 7, is broken down into five main divisions or subjects. Key subjects covered in the NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations are listed below. For a clear knowledge of the ideas associated with this chapter, Extramarks’ experts encourage students to carefully go through each of these topics.

Students will undoubtedly profit fully from Extramarks’ Solutions for this chapter on Simple Equations if they practice the following topics:

• Variable
• An Equation
• Solving an Equation
• Forming an Equation
• Application of Simple Equations to Practical Situations

Students can learn the key information and fundamental equational ideas by using study materials such as NCERT Solutions for Class 7 Maths Chapter 4.The Extramarks platform makes it simple for students to access this content, and they can review and practise more problems in this chapter.

Students should use the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 of problems since, if understood, the fundamentals of algebra can be useful in both academic and practical settings. The number of practise questions for each exercise is listed below for the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 based on Simple Equations:

• Class 7 Maths Chapter 4 Ex 4.1 – 6 Questions
• Class 7 Maths Chapter 4 Ex 4.2 – 4 Questions
• Class 7 Maths Chapter 4 Ex 4.3 – 4 Questions
• Class 7 Maths Chapter 4 Ex 4.4 – 4 Questions

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 begins by explaining what a variable is, how to structure algebraic statements around it, and what an equation is. The topics covered in NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 go into further detail on how to solve equations, perform arithmetic operations within equations, understand the concept of transposition, and create simple algebraic expressions from realistic, everyday situations.

There are a total of 18 questions in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2, 12 of which are simple, 4 of which are fairly difficult, and 2 of which require some conceptual thinking on the part of the students.

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 contain several straightforward yet fundamental ideas for solving equations. The students need to comprehend that even when the left and right sides of an equation are switched, the equation’s balance remains the same. The following are some key ideas addressed in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.2 :

• If either side of the equation is switched from right to left or vice versa, the equation does not change. Therefore, y + 2 = x + 5 and x + 5 = y + 2 are equivalent.
• Moving to the other side is referred to as transposing. When a number is brought to the other side, the sign changes as a result of its transposition.

### NCERT Solutions for Class 7

Students may easily obtain the NCERT Solutions at Extramarks with just a mouse click. Students find it much simpler to prepare for tests when study resources are easily accessible.

The quality of the study materials is effectively upheld by NCERT Solutions. The NCERT Solutions were created by Extramarks professionals with the precise needs of Class 7 students in mind. They include both subjective and MCQ-type questions, as well as their answers and solutions.

Additionally, students have the choice of studying a specific chapter, sitting for a chapter exam, or reviewing the chapter with the NCERT Class 7 Solutions. This enables them to cover every subject in detail and identify any subjects in which they fall short.

The Class 7 Mathematics NCERT Solutions include answers and analyses for all sums from the NCERT textbook that use Triangles, Area, and Perimeter of various shapes, Data management, Integers, etc. Mathematics experts created these NCERT Solutions for Class 7 to accommodate students of various cognitive levels. Assisting CBSE students in developing their Mathematics knowledge and auxiliary abilities.

Mathematics Class 7 NCERT Solutions are very beneficial for students preparing for both internal and external examinations. These answers give students the chance to deepen their understanding of Mathematics in Class 7. Students from other boards who desire solid Mathematical principles can also refer to the CBSE NCERT Solutions for Class 7 Mathematics.

Q.1

$\begin{array}{l}\mathrm{Give}\mathrm{first}\mathrm{the}\mathrm{step}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\\ \mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ \left(\mathrm{a}\right)\mathrm{x}–1=0\left(\mathrm{b}\right)\mathrm{x}+1=0\left(\mathrm{c}\right)\mathrm{x}–1=5\left(\mathrm{d}\right)\mathrm{x}+6=2\\ \left(\mathrm{e}\right)\mathrm{y}–4=–7\left(\mathrm{f}\right)\mathrm{y}–4=4\left(\mathrm{g}\right)\mathrm{y}+4=4\left(\mathrm{h}\right)\mathrm{y}+4=–4\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)x-\text{1}=0\\ \text{Add 1 to both sides to get}\\ \overline{)x=1}\\ \left(\text{b}\right)\text{x}+\text{1}=0\\ \text{Subtract 1 from both sides to get}\\ \overline{)\text{x}=-\text{1}}\\ \left(\text{c}\right)\text{x}-\text{1}=\text{5}\\ \text{Add 1 to both sides to get}\\ \overline{)x=6}\\ \left(\text{d}\right)\text{x}+\text{6}=\text{2}\\ \text{Subtract 6 from both sides to get}\\ \overline{)\text{x}=-\text{4}}\\ \left(\text{e}\right)\text{y}-\text{4}=-\text{7}\\ \text{Add 4 to both sides to get}\\ \overline{)\text{y}=-3}\\ \left(\text{f}\right)\text{y}-\text{4}=\text{4}\\ \text{Add 4 to both sides to get}\\ \overline{)y=0}\\ \left(\text{g}\right)\text{y}+\text{4}=\text{4}\\ \text{Subtract 4 from both sides to get}\\ \overline{)y=0}\\ \left(\text{h}\right)\text{y}+\text{4}=-\text{4}\\ \text{Subtract 4 from both sides to get}\\ \overline{)y=-8}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Give}\mathrm{first}\mathrm{the}\mathrm{step}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\\ \mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ \left(\mathrm{a}\right)3\mathrm{l}=42\left(\mathrm{b}\right)\frac{\mathrm{b}}{2}=6\left(\mathrm{c}\right)\frac{\mathrm{p}}{7}=4\left(\mathrm{d}\right)4\mathrm{x}=25\\ \left(\mathrm{e}\right)8\mathrm{y}=36\left(\mathrm{f}\right)\frac{\mathrm{z}}{3}=\frac{5}{4}\left(\mathrm{g}\right)\frac{\mathrm{a}}{5}=\frac{7}{15}\left(\mathrm{h}\right)2\mathrm{t}=–10\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)3l=42\\ \text{Divide both sides by 3 to get}\\ l=\frac{42}{3}=\overline{)14}\\ \left(\text{b}\right)\frac{b}{2}=6\\ \text{Multiply both sides by 3 to get}\\ \overline{)b=12}\\ \left(\text{c}\right)\frac{p}{7}=\text{4}\\ \text{Multiply both sides by 7 to get}\\ \overline{)p=28}\\ \left(\text{d}\right)\text{4x}=\text{25}\\ \text{Divide both sides by 4 to get}\\ \overline{)x=\frac{25}{4}}\\ \left(\text{e}\right)\text{8y}=36\\ \text{Divide both sides by 8 to get}\\ y=\frac{36}{8}=\overline{)\frac{9}{2}}\\ \left(\text{f}\right)\frac{z}{3}=\frac{5}{4}\\ \text{Multiply both sides by 3 to get}\\ \overline{)z=\frac{15}{4}}\\ \left(\text{g}\right)\frac{a}{5}=\frac{7}{15}\\ \text{Multiply both sides by 5 to get}\\ \text{a=}\frac{7×5}{15}=\frac{7×\overline{)5}}{3×\overline{)5}}=\overline{)\frac{7}{3}}\\ \left(\text{h}\right)2t=-\text{10}\\ \text{Divide both sides by 2 to get}\\ \text{t=}\frac{-10}{2}=\frac{-5×\overline{)2}}{\overline{)2}}=\overline{)-5}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Give}\mathrm{the}\mathrm{steps}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\mathrm{and}\\ \mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ \left(\mathrm{a}\right) 3\mathrm{n}-2 = 46\left(\mathrm{b}\right) 5\mathrm{m}+7 = 17\\ \left(\mathrm{c}\right)\frac{20\mathrm{p}}{\mathrm{3}}= 40\left(\mathrm{d}\right)â€„\frac{3\mathrm{p}}{\mathrm{10}}\mathrm{= 6}\end{array}$

Ans.

$\begin{array}{l}\text{(a)}3n-2=46\\ \text{Add 2 to both sides to get}\\ 3n=48\\ \text{Now, divide both sides by 3 to get}\\ \overline{)n=16}\\ \text{(b)}5m+7=17\\ \text{Subtract 7 from both sides to get}\\ 5m=10\\ \text{Now, divide both sides by 5 to get}\\ \overline{)m=2}\\ \text{(c)}\frac{20p}{3}=40\\ \text{Multiply both sides by 3 to get}\\ 20p=120\\ \text{Now divide both sides by 20, to get}\\ p=\frac{120}{20}=\frac{\overline{)20}×3}{\overline{)20}}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=3}\\ \text{(d)}\text{\hspace{0.17em}}\frac{3p}{10}=6\\ \text{Multiply both sides by 10 to get}\\ 3p=60\\ \text{Now divide both sides by 3, to get}\\ p=\frac{60}{3}=\frac{20×\overline{)3}}{\overline{)3}}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=20}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)10\mathrm{p}=100\left(\mathrm{b}\right)10\mathrm{p}+10=100\left(\mathrm{c}\right)\frac{\mathrm{p}}{4}=5\\ \left(\mathrm{d}\right)\frac{\mathrm{p}}{3}=5\left(\mathrm{e}\right)\frac{3\mathrm{p}}{4}=6\left(\mathrm{f}\right)\mathrm{ }3\mathrm{s}=–9\\ \\ \left(\mathrm{g}\right)\mathrm{ }3\mathrm{s}+12=0\left(\mathrm{h}\right)\mathrm{ }3\mathrm{s}=0\left(\mathrm{i}\right)\mathrm{ }2\mathrm{q}=6\\ \\ \left(\mathrm{j}\right)\mathrm{ }2\mathrm{q}–6=0\left(\mathrm{k}\right)\mathrm{ }2\mathrm{q}+6=0\left(\mathrm{l}\right)\mathrm{ }2\mathrm{q}+6=12\end{array}$

Ans.

$\begin{array}{l}\left(a\right)\text{}10p=100\\ p=\frac{100}{10}=\overline{)10}\\ \text{(b)}10p+10=100\\ 10p=90\\ p=\frac{90}{10}=\overline{)9}\\ \text{(c)}\frac{p}{4}=5\\ p=5×4=\overline{)20}\\ \text{(d)}\frac{p}{3}=5\\ p=5×3=\overline{)15}\\ \left(e\right)\text{}\frac{3p}{4}=6\\ 3p=6×4=24\\ p=\frac{24}{3}=\overline{)8}\\ \left(f\right)\text{\hspace{0.17em}}3s=-9\\ s=\frac{-9}{3}=\overline{)-3}\\ \left(g\right)\text{\hspace{0.17em}}3s+12=0\\ 3s=-12\\ s=\frac{-12}{3}=\overline{)-4}\\ \left(h\right)\text{\hspace{0.17em}}3s=0\\ s=\frac{0}{3}=\overline{)0}\\ \left(i\right)\text{\hspace{0.17em}}2q=6\\ q=\frac{6}{2}=\overline{)3}\\ \left(j\right)\text{\hspace{0.17em}}2q-6=0\\ 2q=6\\ q=\frac{6}{2}=\overline{)3}\\ \left(k\right)\text{\hspace{0.17em}}2q+6=0\\ 2q=-6\\ q=\frac{-6}{2}=\overline{)-3}\\ \left(l\right)\text{\hspace{0.17em}}2q+6=12\\ 2q=12-6\\ 2q=6\\ q=\frac{6}{3}=\overline{)2}\end{array}$