# NCERT Solutions For Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.3) Exercise 2.3

An equation is a mathematical statement with an equal sign between two expressions with equal values. For example, 3x + 5 = 15. There are different types of equations, such as first-order, second-order, and third-order. Students could learn more about Mathematics Equations. An equation is a mathematical formula that contains two algebraic expressions on either side of an equal sign (=). It shows the equivalence relation between the formula written on the left and the formula written on the right. In any formula, L.H.S = R.H.S (left side = right side). Students can solve equations to find the values ​​of unknown variables that represent unknown quantities. If a statement does not have an equal sign, it is not an equation. The difference between expressions and formulas is explained in a later section. For a thorough understanding of linear equations in one variable, students should consult the NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3. These solutions could help them understand the meaning of equations in Mathematics.

Equations have various parts, including coefficients, variables, operators, constants, terms, expressions, and equal signs. When writing an equation, students need the “=” symbol and the term on both sides. Both sides must be equal to each other. The equation does not have to have multiple terms on either side, using variables and operators. Students can also form equations without these. In contrast, equations involving variables are referred to as Algebraic Equations.

Solving a mathematical equation

The equation is like a pair of scales with equal weights on each side. This applies even if students add or subtract the same number on both sides of the equation. The rule applies even if both sides of the equation are multiplied or divided by the same number. To solve basic linear equations using one variable:

Step 1: Students should put all terms, including variables, on one side of the equation and all constants on the other side by doing arithmetic on both sides.

Step 2: Further, students should add, subtract, and combine all equal terms (terms containing the same variable with the same exponent).

Step 3: Students should finally simplify to get the answer. It would be enriching to look at yet another example of the basic equation: 3x – 20 = 7. To bring all the constants to the right side, students would need to add 20 to both sides. This means 3x – 20 + 20 = 7 + 20, which simplifies to 3x = 27. Now, students should divide both sides by 3. This gives x = 9, which is the desired solution of the equation.

Based on the order, Equations can be classified into three types. Below are three types of equations in mathematics.

• Linear Equation
• Cubic Equation

An equation of degree 1 is called a Linear Equation in mathematics. In such an equation, 1 is the highest exponent of the term. These can be further categorized as Linear Equations in One Variable, Linear Equations in Two Variables, Linear Equations in Three Variables and so on. The standard form of a Linear Equation with Two Variables x and y is ax + by – c = 0. b is the coefficient of x or y, and c is a constant.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.3) Exercise 2.3

Prior examination preparation is bound to be advantageous for students. Problems need to be solved in a simplified way so students can easily understand the solutions.

• A note about linear equations

A Linear Equation is defined as an algebraic equation containing Constants and Variables. It is an equation where the left expression has the same value as the right expression. A Linear Equation in One Variable is easily understood as an equation with only one variable. Example: px+q = 0, p and q are two integers, and x is a variable.

The equations studied in junior classes are Linear Equations in One Variable because the terms in such equations have only one variable. These equations are linear because the variables in the equations have a maximum power of 1. Any linear equation can have any rational number as its solution. This concept has been explained descriptively in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 available on the website of Extramarks.

Students can download PDF versions of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, which are compiled by experienced teachers according to the NCERT guidelines. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, consist of problems with comprehensive solutions that can help students revise their entire syllabus and achieve excellence in exams. Students can register on Extramarks to obtain NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, conveniently.

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### Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable

Choosing NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, is considered to be the best option for CBSE students when it comes to exam preparation. With the aid and assistance of the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can master the theme of Linear Equations in One Variable. Students can download these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 at their convenience or study directly online via the Extramarks website or mobile application.

Extramarks’ subject matter experts have engineered the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, meticulously and in compliance with all CBSE guidelines. A Class 8 student who has a thorough understanding of all concepts implicit in the prescribed Mathematics textbook and a good grasp of all the problems from the exercises it contains will easily achieve the best possible score on the final exam. With the help of these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can easily understand the various patterns of questions that may appear on exams pertaining to this chapter and also learn how to study this chapter.

Students can download the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, from the Extramarks website and get ready for their exam. If students have the Extramarks Learning App installed on their smartphone, they can also download NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, from the application. The best part about NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, is that they are accessible both online and offline.

### NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (Ex 2.3)

In addition to these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, which pertain to Exercise 2.3, there are many exercises in this chapter with numerous problems. The solutions to all these exercises have been compiled by highly knowledgeable mentors at Extramarks. Therefore, they should all be of the highest quality and can be referenced by students while preparing for the exam. Understanding all the concepts in the textbook and solving NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 and the adjacent exercise problems is very important to getting the best score in the class.

### NCERT Solutions Class 8 Maths Chapter 2 All Exercises

The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 based on Linear Equations in One Variable, are available in PDF format for download. The NCERT solutions for the Linear Equations in One Variable chapter have been precisely designed by subject matter specialists of Extramarks. It is highly recommended that students ought to refer to the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 to prepare adequately for the exam. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 can also serve as a reference for students’ homework and assignments. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, contain systematic answers for all questions in Exercise 2.3, making them a great learning resource for Class 8 students. By referring to these NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students will be able to perform well on their final exams and master the concepts. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, are developed based on the latest NCERT syllabus and cover all the key topics embedded in Exercise 2.3.

### 12 Questions with Solutions

The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 comprise a continuation of the explanation of concepts of algebraic expressions and equations that students learned in earlier classes. An equation with one variable will be presented to the student in this chapter. Some of the main topics or concepts covered in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 are:

Algebraic equations are equations with variables. The word “equation” indicates that the value of the expression on one side of the equal sign is equal to the value of the expression on the other side. An equation may have, on both sides, linear expressions. Like numbers, variables can be transposed from one side of the equation to the other. Sometimes it is necessary to simplify the formulas that make up an equation before solving it in a general way. Some equations may not even be linear but can be made linear by multiplying both sides of the equation by a suitable formula. The usefulness of linear equations lies in their multiple applications. A variety of problems related to numbers, ages, sizes, combinations of bills, etc. are solved with linear equations.

### 10 Questions with Solutions

The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, focuses on Linear Equations in One Variable are based on solving equations with variables on both sides. The equation states that the values ​​of the two expressions are equal. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, shows how to solve equations that have expressions with variables on both sides.

### NCERT Solutions for Class 8

The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3 provides answers to all questions in Exercise 2.3 of Chapter 2 of the NCERT Class 8 Mathematics textbook. Many questions in the exam are derived from this exercise, and by solving NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.3, students can get a high score in the annual exam.

Q.1

$\begin{array}{l}\text{Solve the following equations and check your results.}\\ 1.\text{ }3\text{x}=2\text{x}+18\\ 2.\text{ }5\text{t}-3=3\text{t}-5\\ 3.\text{ }5\text{x}+9=5+3\text{x}\\ 4.\text{ }4\text{z}+3=6+2\text{z}\\ 5.\text{ }2\text{x}-1=14-\text{x}\\ 6.\text{ }8\text{x}+4=3\left(\text{x}-1\right)+7\\ 7.\text{ x}=\frac{4}{5}\left(\text{x}+10\right)\\ 8.\text{ }\frac{2\text{x}}{3}+1=\frac{7\text{x}}{15}+3\\ 9.\text{ }2\text{y}+\frac{5}{3}=\frac{26}{3}-\text{y}\\ 10.\text{ }3\text{m}=5\text{m}-\frac{8}{5}\end{array}$

Ans

$\begin{array}{l}1.3\mathrm{x}=2\mathrm{x}+18\\ \text{On transposing 2x to L.H.S,}\mathrm{we}\mathrm{get}\\ ⇒3\mathrm{x}-2\mathrm{x}=18\\ ⇒\mathrm{x}=18\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{x}\\ =3×18\\ =54\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=2\mathrm{x}+18\\ =2×18+18\\ =36+18=54\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$

$\begin{array}{l}2.5\mathrm{t}-3=3\mathrm{t}-5\\ \text{On transposing 3t to L.H.S and}-\text{3 to R.H.S,}\text{we get}\\ ⇒5\mathrm{t}-3\mathrm{t}=3-5\\ ⇒2\mathrm{t}=-2\\ ⇒\mathrm{t}=-1\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{t}-3\\ =5×\left(-1\right)-3\\ =-5-3\\ =-8\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3\mathrm{t}-5\\ =3×\left(-1\right)-5\\ =-3-5\\ =-8\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}3.5\mathrm{x}+9=5+3\mathrm{x}\\ \text{On transposing 3x to L.H.S and}9\text{to R.H.S,}\text{we get}\\ ⇒5\mathrm{x}-3\mathrm{x}=5-9\\ ⇒2\mathrm{x}=-4\\ \text{On dividing both sides by 2,we get}\\ ⇒\frac{2\mathrm{x}}{2}=-\frac{4}{2}\\ ⇒\mathrm{x}=-2\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{x}+9\\ =5×\left(-2\right)+9\\ =\left(-10\right)+9\\ =-1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5+3\mathrm{x}\\ =5+3×\left(-2\right)\\ =5+\left(-6\right)\\ =-1\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}4.4\mathrm{z}+3=6+2\mathrm{z}\\ \text{On transposing 2z to L.H.S and}3\text{to R.H.S,}\text{we get}\\ ⇒4\mathrm{z}-2\mathrm{z}=6-3\\ ⇒2\mathrm{z}=3\\ \text{On dividing both sides by 2,}\text{we get}\\ ⇒\frac{2\mathrm{z}}{2}=\frac{3}{2}\\ ⇒\mathrm{z}=\frac{3}{2}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=4\mathrm{z}+3\\ =4×\left(\frac{3}{2}\right)+3\\ =6+3\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=6+2\mathrm{z}\\ =6+2×\frac{3}{2}\\ =6+3\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}5.\mathrm{}2\mathrm{x}-1=14-\mathrm{x}\\ \text{On transposing x to L.H.S and}1\text{to R.H}.\mathrm{S},\mathrm{we}\mathrm{get}\\ ⇒2\mathrm{x}+\mathrm{x}=14+1\\ ⇒3\mathrm{x}=15\\ \text{On dividing both sides by 3,we get}\\ ⇒\frac{3\mathrm{x}}{3}=\frac{15}{3}\\ ⇒\mathrm{x}=5\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{x}-1\\ =2×5-1\\ =10-1\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=14-\mathrm{x}\\ =14-5\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}6.8\mathrm{x}+4=3\left(\mathrm{x}-1\right)+7\\ ⇒8\mathrm{x}+4=3\mathrm{x}-3+7\\ ⇒8\mathrm{x}+4=3\mathrm{x}+4\\ \text{On transposing 3x to L.H.S and}4\text{to R.H.S,we get}\\ ⇒8\mathrm{x}-3\mathrm{x}=4-4\\ ⇒5\mathrm{x}=0\\ ⇒\mathrm{x}=0\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=8\mathrm{x}+4\\ =8×0+4\\ =4\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3\left(\mathrm{x}-1\right)+7\\ =3\left(0-1\right)+7\\ =-3+7\\ =4\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}7.\mathrm{}\mathrm{x}=\frac{4}{5}\left(\mathrm{x}+10\right)\\ \mathrm{Multiplying}\text{}\mathrm{both}\text{}\mathrm{sides}\text{}\mathrm{by}\text{}5,\mathrm{we}\text{}\mathrm{get}\\ ⇒5\mathrm{x}=4\left(\mathrm{x}+10\right)\\ ⇒5\mathrm{x}=4\mathrm{x}+40\\ \mathrm{Transposing}\text{}4\mathrm{x}\text{}\mathrm{to}\text{}\mathrm{L}.\mathrm{H}.\mathrm{S},\mathrm{we}\text{}\mathrm{get}\\ ⇒5\mathrm{x}-4\mathrm{x}=40\\ ⇒\mathrm{x}=40\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{x}\\ =40\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{4}{5}\left(\mathrm{x}+10\right)\\ =\frac{4}{5}\left(40+10\right)\\ =\frac{4}{5}\left(50\right)\\ =\frac{4}{5}×\left(50\right)\\ =4×10\right]\\ =40\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}8.\frac{2\mathrm{x}}{3}+1=\frac{7\mathrm{x}}{15}+3\\ \text{On transposing}\frac{7\mathrm{x}}{15}\text{to L.H.S and}1\text{to R.H.S,we get}\\ ⇒\frac{2\mathrm{x}}{3}-\frac{7\mathrm{x}}{15}=3-1\\ \text{Taking LCM of 3 and 15,we get}\\ ⇒\frac{10\mathrm{x}-7\mathrm{x}}{15}=2\\ ⇒\frac{3\mathrm{x}}{15}=2\\ ⇒3\mathrm{x}=30\\ ⇒\mathrm{x}=\frac{30}{3}=10\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{2\mathrm{x}}{3}+1\\ =\frac{2×10}{3}+1\\ =\frac{20}{3}+1\\ =\frac{20+3}{3}=\frac{23}{3}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{7\mathrm{x}}{15}+3\\ =\frac{7\mathrm{x}+45}{15}\\ =\frac{7×10+45}{15}\\ =\frac{115}{15}\\ =\frac{23}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}9.2\mathrm{y}+\frac{5}{3}=\frac{26}{3}-\mathrm{y}\\ \text{On transposing}\mathrm{y}\text{to L.H.S and}\frac{5}{3}\text{to R.H.S,we get}\\ ⇒2\mathrm{y}+\mathrm{y}=\frac{26}{3}-\frac{5}{3}\\ ⇒3\mathrm{y}=\frac{21}{3}\\ ⇒\mathrm{y}=\frac{21}{9}\\ ⇒\mathrm{y}=\frac{7}{3}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{y}+\frac{5}{3}\\ =2×\frac{7}{3}+\frac{5}{3}\\ =\frac{14}{3}+\frac{5}{3}\\ =\frac{14+5}{3}=\frac{19}{3}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{26}{3}-\mathrm{y}\\ =\frac{26}{3}-\frac{7}{3}\\ =\frac{26-7}{3}\\ =\frac{19}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}10.3\mathrm{m}=5\mathrm{m}-\frac{8}{5}\\ \text{On transposing}5\mathrm{m}\text{to L.H.S ,we get}\\ ⇒3\mathrm{m}-5\mathrm{m}=-\frac{8}{5}\\ ⇒-2\mathrm{m}=-\frac{8}{5}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}⇒\mathrm{m}=-\frac{8}{5×\left(-2\right)}\\ ⇒\mathrm{m}=\frac{4}{5}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{m}\\ =3×\frac{4}{5}\\ =\frac{12}{5}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5\mathrm{m}-\frac{8}{5}\\ =5×\frac{4}{5}-\frac{8}{5}\\ =4-\frac{8}{5}\\ =\frac{20-8}{5}\\ =\frac{12}{5}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ $\begin{array}{l}\\ \end{array}$