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RS Aggarwal Solutions

RS Aggarwal Class 8 Mathematics Solutions for Chapter-8 Linear Equations

RS Aggarwal Class 8 Mathematics Solutions for Chapter-8 Linear Equations

Q: 6. What is a common mistake students make when solving equations with brackets, like 3(x - 2) = 15?

A very common mistake is related to the distributive property. Many students incorrectly multiply the number outside the bracket with only the first term inside, for example, writing 3x - 2 = 15. The correct method requires you to multiply the outer number (3) with every term inside the bracket. The correct first step is 3 * x - 3 * 2 = 15, which simplifies to 3x - 6 = 15. Forgetting to multiply all terms is a frequent source of error.

Q: 8. What is the fundamental difference between an 'expression' and an 'equation' in this chapter?

The fundamental difference lies in the presence of an equality sign (=).An algebraic expression is a combination of variables and constants connected by mathematical operators, but it does not have an equals sign. For example, 5x + 9 is an expression. You can only simplify or evaluate it.An equation, on the other hand, states that two expressions are equal. For example, 5x + 9 = 24 is an equation. The goal with an equation is to 'solve' it—to find the specific value of the variable that makes the statement true.

Class 8 RS Aggarwal Mathematics Linear Equations solutions - free PDF download

Mathematics is a high-scoring subject in which pupils achieve the highest possible score on the exam. When it comes to test preparation, it is the most difficult moment for most students, as they struggle to solve problems. As a result, our skilled teachers at Vedantu created helpful aids to assist students in easily preparing the RS Aggarwal Class 8 Mathematics Solutions for Chapter-8 Linear Equations for their exams. All of the answers are designed and adhere to the most recent CBSE patterns. When students learn on a regular basis, they might acquire simple tactics and shortcut ways. This Chapter's PDFs are available here, and students can download them for free using the link on Vedantu’s site.

Solutions to Linear Equations for Class 8 RS Aggarwal

Scoring good marks in the math final examination becomes easy with the help of RS Aggarwal Class 8 Mathematics Solution Chapter 8. It is recommended that students should practice all the sums given in every math Chapter to score well in the examination. They can download Chapter 8 Class 8 Mathematics RS Aggarwal Solutions PDF from Vedantu. Students can refer to the RS Aggarwal Mathematics Class 8 Chapter 8 Solutions,to clear all their doubts from the Chapter. You can download the RS Aggarwal Class 8 Mathematics Chapter 8 solutions PDF from Vedantu for free. You may also be able to download NCERT Solution and NCERT Solutions for Class 8 Mathematics. Subjects like Science, Mathematics, English will become easy to study if you have access to NCERT Solution for Class 8 Science, Mathematics solutions and solutions of other subjects.

Exercise-Wise Solutions

We have provided step by step solutions for all exercise questions given in the PDF of Class 8 RS Aggarwal Chapter-8 Linear Equations. All the Exercise questions with solutions in Chapter-8 Linear Equations are given below:

Learn More About Linear Equations:

When it comes to the topic of Linear Equations, then there are some concepts that students should be familiar with before they start solving sums. If you want to learn those topics, then you can refer to the thoroughly discussed topics below. All answers given in the RS Aggarwal Solutions for Class 8 Mathematics Chapter 8 PDF file are accurate and reliable.

According to experts, Linear Equations can be defined as equations of the first order. This means that equations are defined for lines in any particular coordinate system. In other words, a Linear Equation is an equation for a straight line.

Usually, the general representation of the straight-line equation can be written as y = mx + b. In this equation, m is the slope of the line and b is the y-intercept. Also, Linear Equations can have the highest exponent value of 1. Some examples of Linear Equations are:

3x - y + z = 4
x + y = 4
x/2 = 7
M + 1 = 0
4y = 8
2x - 3 = 0

Similarly, when an equation has a homogenous variable, which can also mean that there is only one variable, then this type of equation is known as a Linear Equation in one variable. This further means that a line equation can be achieved by relating zero to a linear polynomial over any particular field. This can be used to obtain the coefficients.

If a student finds solutions to Linear Equations, then that will result in the generation of values. These values when substituted for the unknown values would make the equation true. When it comes to the case of one variable, then there is also a single solution.

However, in the case of Linear Equations with two variables, the solutions can be calculated as the Cartesian coordinates of a point of the Euclidean plane. It should be noted here that Linear Equations can be present in one variable, two variables, and three variables. The examples for all of these Linear Equations are:

Linear Equation in One Variable: 3x + 5 = 0, 97x = 47, and 32x + 8 = 0.
Linear Equation in Two Variables: 6x + 9y - 11 = 0, y + 7x = 3, and 3a + 5t = 2.
Linear Equation in Three Variables: 3x + 12y = 2z, x + y + t = 0, and a - 4b = z.

The equation of a straight line can be given by:

Y = mx + b

Here, m is the slope of the line, b is the y-intercept, and x and y are the coordinates of the x-axis and the y-axis, respectively.

Further, if a straight line is parallel to the x-axis, then the x coordinate should be equal to zero. Hence, this means that:

Y = b

Also, if the line is parallel to the y-axis, then the y coordinate will be zero. This means that:

Mx + b = 0

X = -b/m

The slope of the line can also be defined as being equal to the ratio of change in y coordinates and the change in x coordinates. An individual can evaluate the value of a slope by using the following formula.

M = (y2 - y1) / (x2 - x1)

From this formula, it is clear that the slop shows the rise of a line in the plane. This is seen along with the distance covered on the x-axis. Students should remember that the slope of a line is also known as a gradient.

There are other important formulas related to Linear Equations that students should be familiar with. To help students, we have created a table of all those formulas. And that table is mentioned below.

Linear Equation	General Form	Example
Slope intercept form	Y = mx + c	Y + 2x = 4
Point Slope Form	Y - y1 = m (x - x1)	Y - 3 = 6 (x - 2)
General Form	Ax + By + C = 0	2x + 4y - 5 = 0
Intercept Form	x / x0 + y / y0 = 1	x / 2 + y / 4 = 1
As a Function	F (x) instead of y F (x) = x + c	F (x) = x + 4
The Identity Function	F (x) = x	F (x) = 4x
Constant Functions	F (x) = C	F (x) = 7

In this table, m is the slope of a line, and xo and yo are the intercepts of the x-axis and y-axis, respectively.

There are several methods that one should follow to solve Linear Equations in one variable. As a general rule, both sides of the equation should be balanced for solving a Linear Equation. After that, students should solve the equation by using several mathematical operations on both sides of the equation. This should be done in a manner that does not have any effect on the balance of the equation. For example, if one has to solve the equation (2x - 10) / 2 = 3 (x - 1), then this can be done by following the steps that are mentioned below.

Step 1: Begin by clearing the fraction.

X - 5 = 3 (x - 1)

Step 2: Simplify both sides of the equation.

2x - 5 = 3x - 3

X = 3x + 2

X - 3x = 2

Step 3: Isolate the value of x.

-2x = 2

X = -1

In the case of Linear Equations with two variables, then there are different methods in which those equations can be solved. And those ways are:

Method of substitution.
Determinant methods.
Method of elimination.
Cross multiplication method.

Also, when it comes to solving Linear Equations in three variables, then those equations can be solved by using the matrix method. Students will get to learn about those methods in other Classes in the future.

Get the Solutions on Vedantu

There are three exercises in RS Aggarwal Class 8 Chapter 8. In all of these exercises, students are tested on their knowledge of Linear Equations. If a student has any doubts while solving these exercises, then the student may take the help of experts at Vedantu. We provide students with solved R S Aggarwal Class 8 Chapter 8 Solutions PDF for free.

Benefits of Downloading Class 8 Mathematics RS Aggarwal Chapter 8 From Vedantu:

There are many benefits that students can get from downloading Class 8 Mathematics R S Aggarwal Chapter 8 from Vedantu. Some of these benefits are listed below.

All answers are 100% accurate and reliable.
Answers are accompanied by an explanation section that can help students to further understand the answer in a better light.
Answers are written by the best subject matter experts in India.

FAQs on RS Aggarwal Class 8 Mathematics Solutions for Chapter-8 Linear Equations

1. How do you solve a typical linear equation from Exercise 8A of RS Aggarwal Class 8 Maths?

To solve a basic linear equation from RS Aggarwal's Chapter 8, where the variable is on one side, you should follow these steps:

Isolate the variable: The main goal is to get the variable (like 'x') by itself on one side of the equation.
Use transposition: Move the constant terms to the other side of the 'equals' sign. When you move a term, its sign changes (e.g., '+' becomes '-', and '×' becomes '÷').
Simplify: Perform the necessary arithmetic operations to find the value of the variable. For example, in the equation 2x + 5 = 15, you would first transpose 5 to get 2x = 15 - 5, which simplifies to 2x = 10. Finally, you divide by 2 to get x = 5.

2. What is the correct method for solving equations with variables on both sides, as found in RS Aggarwal Chapter 8?

When solving equations with variables on both sides (e.g., 5x - 3 = 3x + 7), the correct method is to consolidate the terms. The steps are:

Group variable terms: Use the transposition method to bring all terms with the variable to one side of the equation (usually the left-hand side).
Group constant terms: Move all the constant terms (numbers without variables) to the opposite side.
Simplify both sides: Combine the like terms. In our example, 5x - 3x = 7 + 3, which simplifies to 2x = 10.
Solve for the variable: Perform the final division or multiplication to find the value of the variable, which here is x = 5.

3. How should word problems from RS Aggarwal Class 8 Chapter 8 be translated into linear equations?

Translating a word problem into a mathematical equation is a key skill. The correct approach involves these four steps:

Identify the Unknown: Read the problem carefully to understand what you need to find. Assign a variable, like 'x' or 'y', to this unknown quantity.
Formulate Expressions: Break down the sentences in the problem and translate them into mathematical expressions involving the variable. For example, "five more than a number" becomes "x + 5".
Create the Equation: Find the statement of equality in the problem that connects the expressions. For instance, if "five more than a number is 20," the equation becomes x + 5 = 20.
Solve the Equation: Use the standard methods of solving linear equations to find the value of the variable, which will be the answer to the word problem.

4. Why is it important to verify the solution after solving a linear equation in Chapter 8?

Verifying your solution is a crucial final step. The primary reason is to ensure accuracy. By substituting the value you found for the variable back into the original equation, you can check if the Left-Hand Side (LHS) equals the Right-Hand Side (RHS). If they match, your solution is correct. This process helps you catch any calculation errors made during transposition or simplification. For exams, this simple check can be the difference between a right and wrong answer, helping secure full marks.

5. How do you solve linear equations involving fractions in RS Aggarwal Class 8 Maths?

To solve linear equations that contain fractions, the most effective method is to eliminate the denominators first. The step-by-step process is:

Find the Least Common Multiple (LCM) of all the denominators in the equation.
Multiply every term on both sides of the equation by this LCM. This will cancel out all the denominators.
You will be left with a simpler linear equation without any fractions.
Solve this new equation using the standard transposition and simplification methods to find the value of the variable.

6. What is a common mistake students make when solving equations with brackets, like 3(x - 2) = 15?

A very common mistake is related to the distributive property. Many students incorrectly multiply the number outside the bracket with only the first term inside, for example, writing 3x - 2 = 15. The correct method requires you to multiply the outer number (3) with every term inside the bracket. The correct first step is 3 * x - 3 * 2 = 15, which simplifies to 3x - 6 = 15. Forgetting to multiply all terms is a frequent source of error.

7. How do the step-by-step solutions for RS Aggarwal Chapter 8 improve exam performance?

Following detailed, step-by-step solutions for Chapter 8 helps in multiple ways. Firstly, it teaches the systematic approach required to solve problems, which is essential for earning full marks in exams. Secondly, by repeatedly practising the correct methodology, you strengthen your understanding of core algebraic concepts like transposition and simplification. This builds a strong foundation, reduces careless errors, and increases your speed and confidence in solving any linear equation during the exam.

8. What is the fundamental difference between an 'expression' and an 'equation' in this chapter?

The fundamental difference lies in the presence of an equality sign (=).

An algebraic expression is a combination of variables and constants connected by mathematical operators, but it does not have an equals sign. For example, 5x + 9 is an expression. You can only simplify or evaluate it.
An equation, on the other hand, states that two expressions are equal. For example, 5x + 9 = 24 is an equation. The goal with an equation is to 'solve' it—to find the specific value of the variable that makes the statement true.