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RS Aggarwal Solutions Class 10 Chapter 3 - Linear Equations in two variables (Ex 3B) Exercise 3.2

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Download Free PDF of RS Aggarwal Solutions Class 10 Chapter 3 - Linear Equations in two variables (Ex 3B) Exercise 3.2 available on Vedantu

RS Aggarwal Maths Class 10 Linear Equations In Two Variables Ex 3b deals with the knowledge of the different methods to solve Linear Equations. The methods adopted to solve Linear Equations in two variables are clearly explained in detail using suitable examples and problems. On solving RS Aggarwal Class 10 Chapter 3 Maths Exercise 3b, students get a concrete idea of simultaneously solving two Linear Equations numerically. The solution by Vedantu further reinforces the learning with its vivid explanation and step-by-step methods of solving problems. In a nutshell, the concise representation of the RS Aggarwal Solutions Class 10 Maths Chapter 3 Exercise 3b by Vedantu truly enables a student to learn Linear Equations in two variables and score reasonably well in all examinations.


Vedantu is a platform that provides free NCERT Solution and other study materials for students. You can also Download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the simply complete Syllabus and score more marks in your examinations.

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Class 10 Chapter 3 RS Aggarwal Solutions For Linear Equations in two variables (Ex 3B) Exercise 3.2

RS Aggarwal Class 10 Maths Exercise 3b: An Overview with Examples

In RS Aggarwal Solutions Class 10 Maths Ex 3b, the solution for any variable i.e. x or y corresponds to a point on a straight line that can be graphically represented. General form of equations in RS Aggarwal Solutions Maths Class 10 Chapter 3b is given by ax + by + c = 0. A pair of Linear Equations in two variables is represented by two such equations, each containing two different variables. 

The general form of a pair of Linear Equations can be represented as:

 a1x + b1y + c1 = 0 

a2x + b2y + c2 = 0

where a1, a2, b1, b2, c1 and c2 are constants, such that a12 + b12 ≠ 0, a22 + b22 ≠ 0.

In case of a simultaneous pair of Linear Equations mentioned in RS Aggarwal Maths Class 10 Exercise 3b Solutions, the solutions for the variables x and y should satisfy both the equations.


A Pair of Linear Equations in Two Variables can be Represented and Solved by Using the Following Methods:

  • Graphical method: Two lines that represent each linear equation are plotted on the graph and their point of cross-section represents the solution.

  • Algebraic method: The various algebraic methods are:

  • Substitution method

  • Elimination method

  • Cross-multiplication method

RS Aggarwal Class 10th Exercise 3b Maths contains problems based on Linear Equations that use algebraic methods of elimination and substitution.

Example of RS Aggarwal Maths Class 10 Exercise 3b Solutions

Solve for: 3x - y = 23

x/3 + y/4 = 4 

Ans: 3x - y = 23 ……..(i)

x/3 + y/4 = 4 ……………(ii)

On multiplying equation (ii) by 4 we get:-

4x/3 + y = 16 ……………(iii)

Now adding (i) and (iii) we get:-

3x + 4x/3  = 23 + 16

(9x+4x) /3=39 

13x = 117

∴ x = 9

On substituting the value of x in (i)

3 × 9 – y = 23

y = 27 – 23

∴ y = 4

Class 10 Maths RS Aggarwal Solutions Chapter 3 Exercise 3b is curated by our eminent group of educators to ensure better understanding and successful revision. Maths RS Aggarwal Solutions Class 10 Chapter 3 Ex 3b are provided with detailed explanations and solutions that are critical in scoring handsome points in the exams.

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FAQs on RS Aggarwal Solutions Class 10 Chapter 3 - Linear Equations in two variables (Ex 3B) Exercise 3.2

1. What is the correct step-by-step process for solving a problem using the substitution method in RS Aggarwal Class 10 Maths, Chapter 3?

To solve a pair of linear equations from RS Aggarwal Chapter 3 using the substitution method, follow these steps as per the CBSE guidelines for the 2025-26 session:

  • Step 1: Isolate a Variable: Choose one of the two equations and solve it for one variable (either x or y) in terms of the other.

  • Step 2: Substitute: Take the expression for the variable you found in Step 1 and substitute it into the second equation. This will result in an equation with only one variable.

  • Step 3: Solve the Equation: Solve the new single-variable equation to find its value.

  • Step 4: Back-Substitute: Substitute the value found in Step 3 back into the expression from Step 1 to find the value of the other variable.

This provides the complete solution (x, y) for the system.

2. What is the standard form of a pair of linear equations that I should use before solving questions in RS Aggarwal Ex 3B?

Before applying any solving method to questions in RS Aggarwal Chapter 3, it is best practice to write the equations in their standard form. The general form for a pair of linear equations in two variables, x and y, is:

  • a₁x + b₁y + c₁ = 0

  • a₂x + b₂y + c₂ = 0

Here, a₁, b₁, c₁, a₂, b₂, and c₂ are real numbers, and both a and b are not zero in their respective equations. Arranging the equations this way helps in consistently applying methods like substitution and elimination and in checking the conditions for unique, no, or infinite solutions.

3. How can I verify if my final answer (x, y) for an RS Aggarwal Chapter 3 problem is correct?

Verifying your solution is a crucial step to ensure accuracy. To check if your values for x and y are correct, you must substitute them back into both original equations. Your solution is correct only if it satisfies both equations simultaneously, meaning the Left-Hand Side (LHS) equals the Right-Hand Side (RHS) for each one. If the values only work for one of the two equations, there has been a calculation error.

4. What is a common mistake students make when using the substitution method for problems in this chapter?

A very common mistake when using the substitution method is an error in sign handling. When you substitute an expression like (3 - 2x) for y in an equation like 4x - 5y = 10, you must multiply the entire expression by -5. Students often forget to distribute the negative sign correctly, writing 4x - 5(3 - 2x) and then incorrectly calculating it as 4x - 15 - 10x instead of the correct 4x - 15 + 10x. Always use parentheses during substitution to avoid this error.

5. How does the substitution method show whether a system of equations has a unique solution, no solution, or infinitely many solutions?

The outcome of the substitution process reveals the nature of the solution:

  • Unique Solution: The method will yield specific numerical values for both x and y. This happens when the lines represented by the equations intersect at a single point.

  • No Solution: After substitution, the variables will cancel out, leaving a false statement (e.g., 5 = 9). This indicates that the lines are parallel and never intersect.

  • Infinitely Many Solutions: The variables will cancel out, but you will be left with a true statement (e.g., 7 = 7). This means the equations represent the same line (coincident lines), and any point on the line is a solution.

6. What is the most important first step when solving a word problem from RS Aggarwal Chapter 3?

The most critical first step in solving a word problem involving linear equations is to identify the unknown quantities and assign variables to them. Carefully read the problem to determine what you need to find. For example, if the problem involves prices of articles or ages of people, you would start by stating: 'Let the cost of one pen be ₹x and the cost of one notebook be ₹y.' Once the variables are clearly defined, you can translate the information from the problem into two distinct linear equations to be solved.

7. How is the substitution method fundamentally different from the elimination method for solving linear equations?

The fundamental difference lies in their approach to removing a variable:

  • The substitution method works by replacing one variable. You solve one equation for one variable (e.g., y = mx + c) and substitute this entire expression into the other equation.

  • The elimination method works by cancelling one variable. You manipulate one or both equations (by multiplying them by constants) so that the coefficients of one variable are equal and opposite. Then, you add the two equations together, causing that variable to be eliminated.

While both methods yield the same result, one may be more efficient than the other depending on the structure of the given equations.