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# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.2 - Pair of Linear Equations in Two Variables

Last updated date: 17th Jul 2024
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## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2 - FREE PDF Download

Linear Equation is a vital topic in Mathematics. No matter how advanced studies you go for, the linear equation concept will always help you solve critical problems by finding values of unknown variables. The study material of Class 10 Maths Chapter 3 Exercise 3.2 Solutions is a wonderful resource that helps you to understand this chapter in an easy way. It not only explains the basics in a step-by-step manner but also provides a wide array of problems and solutions. By practising these problems, you can surely get a good grasp of the chapter. Vedantu has come up with easy to access Maths NCERT Solutions for Exercise 3.2 Class 10 Chapter 3 pdf. Maths Class 10 NCERT Solutions Chapter 3 is available free!

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 3 Exercise 3.2 Class 10 | Vedantu
3. Access PDF for Maths NCERT Chapter 3 Pair of Linear Equations in Two Variables
3.1Class 10 Ex 3.2
4. Class 10 Maths Chapter 3: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 3 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 3 Exercise 3.2 Class 10 | Vedantu

• NCERT Solutions for Class 10th Maths Chapter 3 Exercise 3.2, focuses on solving pairs of linear equations in two variables using various algebraic methods such as substitution, elimination, and cross-multiplication.

• This exercise deals with equations represented in the general form of ax + by = c, where a, b, and c are constants, and x and y are the variables.

• Solve systems of linear equations using the substitution method.

• Interpreting the solutions obtained, which can be unique solutions, infinitely many solutions, or no solutions.

• This exercise is crucial as it not only sharpens students' problem-solving and analytical skills but also illustrates the practical application of mathematical principles in everyday situations.

• In Class 10th Maths, Chapter 3, Exercise 3.2 Pair of Linear Equations in Two Variables there are 2 Solved Questions.

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## Access PDF for Maths NCERT Chapter 3 Pair of Linear Equations in Two Variables

### Class 10 Ex 3.2

1. Solve the following pair of linear equations by the substitution method.

(i) $x+y=14;x-y=4$

Ans: The given equations are:

$x+y=14$           …… (i)

$x-y=4$             …… (ii)

From equation (i):

$x=14-y$          …… (iii)

Substituting (iii) in equation (ii), we get

$\left( 14-y \right)-y=4$

$14-2y=4$

$10=2y$

$y=5$                  …… (iv)

Substituting (iv) in (iii), we get

$x=9$

Therefore, $x=9$ and $y=5$.

(ii) $s-t=3;\frac{s}{3}+\frac{t}{2}=6$

Ans: The given equations are:

$s-t=3$           …… (i)

$\frac{s}{3}+\frac{t}{2}=6$       …… (ii)

From equation (i):

$s=t+3$          …… (iii)

Substituting (iii) in equation (ii), we get

$\frac{t+3}{3}+\frac{t}{2}=6$

$2t+6+3t=36$

$5t=30$

$t=6$                  …… (iv)

Substituting (iv) in (iii), we get

$s=9$

Therefore, $s=9$ and $t=6$.

(iii) $3x-y=3;9x-3y=9$

Ans: The given equations are:

$3x-y=3$           …… (i)

$9x-3y=9$             …… (ii)

From equation (i):

$y=3x-3$          …… (iii)

Substituting (iii) in equation (ii), we get

$9x-3\left( 3x-3 \right)=9$

$9x-9x+9=9$

$9=9$

For all $x$ and $y$.

Therefore, the given equations have infinite solutions. One of the solution is $x=1,y=0$.

(iv) $0.2x+0.3y=1.3;0.4x+0.5y=2.3$

Ans: The given equations are:

$0.2x+0.3y=1.3$           …… (i)

$0.4x+0.5y=2.3$             …… (ii)

From equation (i):

$x=\frac{1.3-0.3y}{0.2}$          …… (iii)

Substituting (iii) in equation (ii), we get

$0.4\left( \frac{1.3-0.3y}{0.2} \right)+0.5y=2.3$

$2.6-0.6y+0.5y=2.3$

$2.6-2.3=0.1y$

$y=3$                  …… (iv)

Substituting (iv) in (iii), we get

$x=\frac{1.3-0.3\left( 3 \right)}{0.2}$

$x=2$

Therefore, $x=2$ and $y=3$.

(v) $\sqrt{2}x+\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0$

Ans: The given equations are:

$\sqrt{2}x+\sqrt{3}y=0$           …… (i)

$\sqrt{3}x-\sqrt{8}y=0$             …… (ii)

From equation (i):

$x=\frac{-\sqrt{3}y}{\sqrt{2}}$          …… (iii)

Substituting (iii) in equation (ii), we get

$\sqrt{3}\left( \frac{-\sqrt{3}y}{\sqrt{2}} \right)-\sqrt{8}y=0$

$(\dfrac{-3} { \sqrt{2}}) y-\sqrt{8} y=0$

$y=0$                  …… (iv)

Substituting (iv) in (iii), we get

$x=0$

Therefore, $x=0$ and $y=0$.

(vi) $\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Ans: The given equations are:

$\frac{3x}{2}-\frac{5y}{3}=-2$           …… (i)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$             …… (ii)

From equation (i):

$x=\frac{-12+10y}{9}$          …… (iii)

Substituting (iii) in equation (ii), we get

$\frac{\left( \frac{-12+10y}{9} \right)}{3}+\frac{y}{2}=\frac{13}{6}$

$\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}$

$\frac{-24+20y+27y}{54}=\frac{13}{6}$

$47y=141$

$y=3$                  …… (iv)

Substituting (iv) in (iii), we get

$x=2$

Therefore, $x=2$ and $y=3$.

2. Solve $\mathbf{2x}+\mathbf{3y}=\mathbf{11}$ and $\mathbf{2x}-\mathbf{4y}=-\mathbf{24}$ and hence find the value of ‘$m$’ for which $\mathbf{y}=\mathbf{mx}+\mathbf{3}$.

Ans: The given equations are:

$2x+3y=11$           …… (i)

$2x-4y=-24$             …… (ii)

From equation (i):

$x=\frac{11-3y}{2}$          …… (iii)

Substituting (iii) in equation (ii), we get

$2\left( \frac{11-3y}{2} \right)-4y=-24$

$11-3y-4y=-24$

$-7y=-35$

$y=5$                  …… (iv)

Substituting (iv) in (iii), we get

$x=-2$

Therefore, $x=-2$ and $y=5$.

Calculating the value of $m$:

$y=mx+3$

$5=-2m+3$

$m=-1$

3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is $\mathbf{26}$ and one number is three times the other. Find them.

Ans: Assuming one number be $x$ and another number be $y$ such that $y>x$,

Writing the algebraic representation using the information given in the question:

$y=3x$             …… (i)

$y-x=26$    …… (ii)

Substituting the value of $y$ from equation (i) in equation (ii), we get

$3x-x=26$

$2x=26$

$x=13$                  …… (iii)

Substituting (iii) in (i), we get

$y=39$

Therefore, numbers are $13$ & $39$.

(ii) The larger of two supplementary angles exceeds the smaller by $\mathbf{18}$ degrees. Find them.

Ans: Assuming the larger angle be $x$ and smaller angle be $y$.

The sum of a pair of supplementary angles is always ${{180}^{\circ }}$.

Writing the algebraic representation using the information given in the question:

$x+y=180$             …… (i)

$x-y=18$    …… (ii)

Substituting the value of $x$ from equation (i) in equation (ii), we get

$180-y-y=18$

$162=2y$

$y=81$                  …… (iii)

Substituting (iii) in (i), we get

$x=99$

Therefore, the two angles are  $x={{99}^{\circ }}$ and $y={{81}^{\circ }}$.

(iii) The coach of a cricket team buys $\mathbf{7}$ bats and 6 balls for $\mathbf{Rs}\text{ }\mathbf{3800}$. Later, she buys $\mathbf{3}$ bats and $\mathbf{5}$ balls for $\mathbf{Rs}\text{ }\mathbf{1750}$. Find the cost of each bat and each ball.

Ans: Assuming the cost of a bat is $x$ and the cost of a ball is $y$.

Writing the algebraic representation using the information given in the question:

$7x+6y=3800$             …… (i)

$3x+5y=1750$    …… (ii)

From equation (i):

$y=\frac{3800-7x}{6}$       …… (iii)

Substituting (iii) in equation (ii):

$3x+5\left( \frac{3800-7x}{6} \right)=1750$

$3x+\frac{9500}{3}-\frac{35x}{6}=1750$

$3x-\frac{35x}{6}=1750-\frac{9500}{3}$

$\frac{18x-35x}{6}=\frac{5250-9500}{3}$

$\frac{-17x}{6}=\frac{-4250}{3}$

$x=500$                  …… (iv)

Substituting (iv) in (iii), we get

$y=\frac{3800-7\left( 500 \right)}{6}$

$y=50$

Therefore, the bat costs $Rs\text{ }500$ and the ball costs $Rs\text{ }50$.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $\mathbf{10}\text{ }\mathbf{km}$, the charge paid is $\mathbf{Rs}\text{ }\mathbf{105}$ and for a journey of $\mathbf{15}\text{ }\mathbf{km}$, the charge paid is $\mathbf{Rs}\text{ }\mathbf{155}$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $\mathbf{25}\text{ }\mathbf{km}$ ?.

Ans: Assuming the fixed charge be $Rs\text{ }x$ and the per km charge be $Rs\text{ y}$.

Writing the algebraic representation using the information given in the question:

$x+10y=105$             …… (i)

$x+15y=155$    …… (ii)

From equation (i):

$x=105-10y$       …… (iii)

Substituting (iii) in equation (ii):

$105-10y+15y=155$

$5y=50$

$y=10$                  …… (iv)

Substituting (iv) in (iii), we get

$x=105-10\left( 10 \right)$

$x=5$

Therefore, the fixed charge is $Rs\text{ }5$ and the per km charge is $Rs\text{ 10}$.

So, charge for $25\text{ }km$ will be:

$=Rs\text{ }\left( x+25y \right)$

$=Rs\text{ 255}$

(v) A fraction becomes $\frac{9}{11}$, if $\mathbf{2}$ is added to both the numerator and the denominator. If, $\mathbf{3}$ is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.

Ans: Assuming the fraction be $\frac{x}{y}$.

Writing the algebraic representation using the information given in the question:

$\frac{x+2}{y+2}=\frac{9}{11}$

$11x+22=9y+18$

$11x-9y=-4$             …… (i)

$\frac{x+3}{y+3}=\frac{5}{6}$

$6x+18=5y+15$

$6x-5y=-3$    …… (ii)

From equation (i):

$x=\frac{-4+9y}{11}$       …… (iii)

Substituting (iii) in equation (ii):

$6\left( \frac{-4+9y}{11} \right)-5y=-3$

$-24+54y-55y=-33$

$y=9$                  …… (iv)

Substituting (iv) in (iii), we get

$x=\frac{-4+9\left( 9 \right)}{11}$

$x=7$

Therefore, the fraction is $\frac{7}{9}$.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Ans: Assuming the age of Jacob be $x$ and the age of his son be $y$.

Writing the algebraic representation using the information given in the question:

$\left( x+5 \right)=3\left( y+5 \right)$

$x-3y=10$             …… (i)

$\left( x-5 \right)=7\left( y-5 \right)$

$x-7y=-30$    …… (ii)

From equation (i):

$x=3y+10$       …… (iii)

Substituting (iii) in equation (ii):

$3y+10-7y=-30$

$-4y=-40$

$y=10$                  …… (iv)

Substituting (iv) in (iii), we get

$x=3\left( 10 \right)+10$

$x=40$

Therefore, Jacob’s present age is $40$ years and his son’s present age is $10$ years.

## Conclusion

NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2 act as your perfect guide to conquering linear equations with two variables. This exercise is essential for mastering this fundamental concept. The key understanding from class 10th exercise 3.2 is to make use of methods like elimination and substitution to solve these equations and interpret solutions based on how the equations' graph lines intersect, run parallel, or coincide. By diligently practicing these methods with NCERT Solutions for Class 10 Maths Chapter 3.2 Students can score good marks in their exams.

## Class 10 Maths Chapter 3: Exercises Breakdown

 Chapter 3 - Pair of Linear Equations in Two Variables Exercises in PDF Format Exercise 3.1 7 Questions & Solutions Exercise 3.3 2 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.2 - Pair of Linear Equations in Two Variables

1. What are the Methods by Which the Pair of Linear Equations can be Solved in Two Variables in Class 10 Chapter 3 Exercise 3.3?

Three methods can be applied to solve the pair of linear equations in two variables which are cross multiplication method, elimination method and substitution method. Also, the students must solve the three questions given in Maths Exercise 3.3 Class 10. In the cross multiplication method, you get to multiply the numerator of a fraction to the denominator of another fraction and denominator of one fraction is multiplied by the numerator of another fraction.

2. Define Substitution and Elimination Methods in Class 10 Maths Ch 3 Ex 3.3.

The substitution method is used to find the value of one variable. On the other hand, the elimination method is used for getting an equation in one variable by adding or subtracting the equations. If the coefficients of one variable are opposite, then you have to add the equations so that a variable is eliminated. Similarly, if the coefficients are equal, to eliminate the variable, you have to subtract the equation. Students will get to know more about the chapter through Exercise 3.3 Class 10th Maths and also by referring to the solution.

3. How many questions are there in Exercise 3.3 of Class 10 Maths?

Exercise 3.3 of Class 10 Maths consists of three main questions with six sub-questions in the first and the third main question. Students can score well in Maths only by practice. Vedantu provides chapter-wise solutions for Maths that are helpful when you are stuck while solving a problem. One of the best things about Vedantu is that these solutions are accessible anytime for free. The students can download the PDF for referring them offline. Each solution is thorough and effective.

4. How many examples are based on Exercise 3.3 of Class 10 Mathematics?

Chapter 3 is Linear Equations and there are only four examples on which Exercise 3.3 is based for Class 10 Mathematics. Vedantu Solutions are provided to explain the basics in a step-by-step manner and it also highlights the important concepts. For example, in the 3.3 Exercise, the substitution method is very important. However, the subject matter experts have verified each solution thoroughly. The only way to score high marks in the examination is by practice.

5. What are linear equations in two variables class 10 maths ex 3.2?

In NCERT Ex 3.2 Class 10 Linear equations in two variables are algebraic expressions that form a straight line when graphed on a coordinate plane. The constants cannot be zero, as this would not produce a valid linear equation. These equations describe a relationship where every point on the line satisfies the equation. Understanding these equations is fundamental in algebra because they provide a basis for more complex mathematical concepts and are widely used in various applications, from basic geometry to advanced calculus and real-world problem-solving scenarios.

6. What are the topics that are covered in exercises 3.3 of Class 10 Mathematics?

Exercise 3.3 of Class 10 Mathematics covers the topic of the substitution method, elimination method, and cross-multiplication method used for solving the linear equations. The examples given in the NCERT Textbook and the solutions provided by the Vedantu guide offer the step-by-step and detailed answers for all three main questions. The solutions are provided by experienced subject-matter experts and are set by following the guidelines provided by the CBSE Board.

7. How many methods are taught in this exercise for solving equations ex 3.2 class 10?

In NCERT Solutions for Class 10 Chapter 3 Exercise 3.2, two main methods for solving pairs of linear equations in two variables are covered: the substitution method and the elimination method. Here's a detailed explanation of each:

• Substitution Method: The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. This process reduces the system of two equations in two variables to a single equation in one variable, making it simpler to solve.

• Elimination Method: The elimination method works by adding or subtracting the equations in such a way that one of the variables is eliminated. This method is particularly useful when the coefficients of one of the variables are opposites or can easily be made opposites.

8. Are there any graphical representations involved in class 10 maths ch 3 ex 3.2?

In Ex 3.2 Class 10 of Chapter 3 from the NCERT textbook, the focus is primarily on solving pairs of linear equations using algebraic methods—specifically the substitution and elimination methods. This particular exercise does not typically involve graphical representations directly. The aim is to teach students how to handle these equations algebraically, which is a crucial skill for further mathematical studies and standardized tests.