# NCERT Solutions for Class 10 Maths (Chapter 3, Exercise 3.2)

NCERT Solutions for Class 10 Maths Chapter 3 is provided in this section. The Chapter 3 of Class 10 Mathematics under the NCERT Syllabus is about the Pair of Linear Equations in Two Variables. As per the Textbook published by NCERT, the chapter discusses the concepts of Pair of Linear Equations in Two Variables, Graphical Method of Solution of a Pair of Linear Equations, Algebraic Methods of Solving a Pair of Linear Equations, Substitution Method, Elimination Method, Cross-Multiplication Method and Equations Reducible to a Pair of Linear Equations in Two Variables.

This post contains the step-by-step solutions of Exercise 3.2, Chapter 3 of Class 10 Mathematics designed in the easiest possible way. There is a total of 7 questions in the Ex. 3.2 and all the questions are covered in this section.

### NCERT Solutions for Class 10 Maths, Chapter 3 – Pair of Linear Equations in Two Variables (Ex. 3.2)

NCERT Solutions for Class 10 Maths, Chapter 3 (Pair of Linear Equations in Two Variables), Exercise 3.2 are described below. The Exercise 3.2 is based on the Graphical Method of Solution of a Pair of Linear Equations and each and every question of Ex. 3.2 are solved (step-by-step) below.

## Question 1: Form the pair of linear equations in the following problems and find their solutions graphically.

### i. 10 students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, then find the number of boys and girls who took part in the quiz.

Solution:

Let us consider the number of boys be $x$ and the number of girls as $y$.

According to the question, the total number of students who took part in the mathematics quiz is 10.

$\therefore x+y=10\,\,\,\,\,\,\,\,\,\,……(i)$

Also, the number of girls is 4 more than the number of boys.

$\therefore y=x+4\,\,\,\,\,\,\,\,\,\,……(ii)$

From Equation $(i)$

$x+y=10$

$\Rightarrow y=10-x$

Similarly, from Equation $(ii)$

$y=x+4$

Plotting Table 1 and 2 we get,

The two lines intersect at the point $(3, 7)$.

Therefore, $x=3$ and $y=7$ is the required solution of the obtained pair of linear equations.

Hence the required number of boys is $3$ and girls is $7$.

### ii. 5 pencils and 7 pens together cost ₹50 whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.

Solution:

Let,

The cost of one pencil be $₹x$ and

The cost of one pen be $₹y$

According to the questions,

The cost of 5 pencils and 7 pens together is ₹50.

$\therefore 5x+7y=50\,\,\,\,\,\,\,\,\,\,……(i)$

Again,

The cost of 7 pencils and 5 pens together is ₹46.

$\therefore 7x+5y=46\,\,\,\,\,\,\,\,\,\,……(ii)$

From Equation $(i)$

$5x+7y=50$

$\Rightarrow y=\frac{50-5x}{7}$

Similarly, from Equation $(ii)$

$7x+5y=46$

$\Rightarrow y=\frac{46-7x}{5}$

Plotting Table 1 and 2 we get,

The two lines intersect at the point $(3, 5)$.

Therefore, $x=3$ and $y=5$ is the required solution of the obtained pair of linear equations.

Hence the cost of one pencil is $₹3$ and the cost of one pen is $₹5$.

## Question 2: On comparing the ratios $\frac{a_1}{a_2},\,\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pair of linear equations intersect at a point , are parallel or coincident.

### i.$5x-4y+8=0;\,7x+6y-9=0$

Solution:

The given pair of linear equations is,

$5x-4y+8=0\,\,\,\,\,\,\,\,\,\,……(i)$

$7x+6y-9=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=5,\,b_1=-4$ and $c_1=8$

Similarly, from equation $(i)$

$a_2=7,\,b_2=6$ and $c_2=-9$

Now,

$\frac{a_1}{a_2}=\frac{5}{7}$

$\frac{b_1}{b_2}=\frac{-4}{6}$

$\frac{c_1}{c_2}=\frac{8}{-9}$

Here,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, the lines representing the given pair of linear equations intersect at a point.

### ii. $9x+3y+12=0;\,18x+6y+24=0$

Solution:

The given pair of linear equations is,

$9x+3y+12=0\,\,\,\,\,\,\,\,\,\,……(i)$

$18x+6y+24=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=9,\,b_1=3$ and $c_1=12$

Similarly, from equation $(i)$

$a_2=18,\,b_2=6$ and $c_2=24$

Now,

$\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, the lines representing the given pair of linear equations are coincident.

### iii. $6x-3y+10=0;\,2x-y+9=0$

Solution:

The given pair of linear equations is,

$6x-3y+10=0\,\,\,\,\,\,\,\,\,\,……(i)$

$2x-y+9=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=6,\,b_1=-3$ and $c_1=10$

Similarly, from equation $(i)$

$a_2=2,\,b_2=-1$ and $c_2=9$

Now,

$\frac{a_1}{a_2}=\frac{6}{2}=3$

$\frac{b_1}{b_2}=\frac{-3}{-1}=3$

$\frac{c_1}{c_2}=\frac{10}{9}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the lines representing the given pair of linear equations are parallel.

## Question 3: On comparing the ratios $\frac{a_1}{a_2},\,\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the following pairs of linear equations are consistent or inconsistent.

### i. $3x+2y=5;\,2x-3y=7$

Solution:

The given pair of linear equations is,

$3x+2y=5$

$\Rightarrow 3x+2y-5=0\,\,\,\,\,\,\,\,\,\,……(i)$

$2x-3y=7$

$\Rightarrow 2x-3y-7=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=3,\,b_1=2$ and $c_1=-5$

Similarly, from equation $(ii)$

$a_2=2,\,b_2=-3$ and $c_2=-7$

Now,

$\frac{a_1}{a_2}=\frac{3}{2}$

$\frac{b_1}{b_2}=\frac{2}{-3}$

$\frac{c_1}{c_2}=\frac{5}{7}$

Here,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, the given pair of linear equations are consistent.

### ii.$2x-3y=8;\,4x-6y=9$

Solution:

The given pair of linear equations is,

$2x-3y=8$

$\Rightarrow 2x-3y-8=0\,\,\,\,\,\,\,\,\,\,……(i)$

$4x-6y=9$

$\Rightarrow 4x-6y-9=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=2,\,b_1=-3$ and $c_1=-8$

Similarly, from equation $(ii)$

$a_2=4,\,b_2=-6$ and $c_2=-9$

Now,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{8}{9}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the given pair of linear equations are inconsistent.

### iii.$\frac{3}{2}x+\frac{5}{3}y=7;\,9x-10y=14$

Solution:

The given pair of linear equations is,

$\frac{3}{2}x+\frac{5}{3}y=7$

$\Rightarrow \frac{9x+10y}{6}=7$

$\Rightarrow 9x+10y=42$

$\Rightarrow 9x+10y-42=0\,\,\,\,\,\,\,\,\,\,……(i)$

$9x-10y=14$

$\Rightarrow 9x-10y-14=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=9,\,b_1=10$ and $c_1=-42$

Similarly, from equation $(ii)$

$a_2=9,\,b_2=-10$ and $c_2=-14$

Now,

$\frac{a_1}{a_2}=\frac{9}{9}=1$

$\frac{b_1}{b_2}=\frac{10}{-10}=-1$

$\frac{c_1}{c_2}=\frac{-42}{-14}=3$

Here,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, the given pair of linear equations are consistent.

### iv.$5x-3y=11;\,-10x+6y=-22$

Solution:

The given pair of linear equations is,

$5x-3y=11$

$\Rightarrow 5x-3y-11=0\,\,\,\,\,\,\,\,\,\,……(i)$

$-10x+6y=-22$

$\Rightarrow -10x+6y+22=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=5,\,b_1=-3$ and $c_1=-11$

Similarly, from equation $(ii)$

$a_2=-10,\,b_2=6$ and $c_2=22$

Now,

$\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, the given pair of linear equations are consistent.

### v.$\frac{4}{3}x+2y=8;\,2x+3y=12$

Solution:

The given pair of linear equations is,

$\frac{4}{3}x+2y=8$

$\Rightarrow \frac{4x+6y}{3}=8$

$\Rightarrow 4x+6y=24$

$\Rightarrow 4x+6y-24=0\,\,\,\,\,\,\,\,\,\,……(i)$

$2x+3y=12$

$\Rightarrow 2x+3y-12=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=4,\,b_1=6$ and $c_1=-24$

Similarly, from equation $(ii)$

$a_2=2,\,b_2=3$ and $c_2=-12$

Now,

$\frac{a_1}{a_2}=\frac{4}{2}=2$

$\frac{b_1}{b_2}=\frac{6}{3}=2$

$\frac{c_1}{c_2}=\frac{-24}{-12}=2$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, the given pair of linear equations are consistent.

## Question 4: Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solutions graphically.

### i.$x+y=5;\,\,2x+2y=10$

Solution:

The given pair of linear equations is,

$x+y=5$

$\Rightarrow x+y-5=0\,\,\,\,\,\,\,\,\,\,……(i)$

$2x+2y=10$

$\Rightarrow 2x+2y-10=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=1,\,b_1=1$ and $c_1=-5$

Similarly, from equation $(i)$

$a_2=2,\,b_2=2$ and $c_2=-10$

Now,

$\frac{a_1}{a_2}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, the given pair of linear equations is consistent.

Again,

If we divide the equation$(ii)$ by $2$ on both the sides, it becomes similar to equation$(i)$. The equations will have same values for $x$ and $y$.

$\therefore x+y=5$

$\Rightarrow y=5-x$

Plotting the values of $x$ and $y$ we get

It has been observed that the graphs of both the equations are similar or coincident with each other. Therefore, the pair of linear equations will have infinitely many solutions.

### ii.$x-y=8;\,\,3x-3y=16$

Solution:

The given pair of linear equations is,

$x-y=3$

$\Rightarrow x-y-3=0\,\,\,\,\,\,\,\,\,\,……(i)$

$3x-3y=16$

$\Rightarrow 3x-3y-16=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=1,\,b_1=-1$ and $c_1=-3$

Similarly, from equation $(i)$

$a_2=3,\,b_2=-3$ and $c_2=-16$

Now,

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}$

$\frac{c_1}{c_2}=\frac{-3}{-16}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the given pair of linear equations are inconsistent.

### iii.$2x+y-6=0;\,\,4x-2y-4=0$

Solution:

The given pair of linear equations is,

$2x+y-6=0\,\,\,\,\,\,\,\,\,\,……(i)$

$4x-2y-4=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=2,\,b_1=1$ and $c_1=-6$

Similarly, from equation $(i)$

$a_2=4,\,b_2=-2$ and $c_2=-4$

Now,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}$

Here,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Therefore, the given pair of linear equations are consistent.

From Equation $(i)$

$2x+y-6=0$

$\Rightarrow y=6-2x$

Similarly, from Equation $(ii)$

$4x-2y-4=0$

$\Rightarrow y=2x-2$

Plotting Table 1 and 2 we get,

The two lines intersect at the point $(2, 2)$.

Therefore, $x=2$ and $y=2$ is the required solution of the given pair of linear equations.

### iv.$2x-2y-2=0;\,\,4x-4y-5=0$

Solution:

The given pair of linear equations is,

$2x-2y-2=0\,\,\,\,\,\,\,\,\,\,……(i)$

$4x-4y-5=0\,\,\,\,\,\,\,\,\,\,……(ii)$

From equation $(i)$

$a_1=2,\,b_1=-2$ and $c_1=-2$

Similarly, from equation $(i)$

$a_2=4,\,b_2=-4$ and $c_2=-5$

Now,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-2}{-5}$

Here,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Therefore, the given pair of linear equations are inconsistent.

## Question 5: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Let,

The length of the garden $=x\,m$ and

The width of the garden $=y\,m$

The perimeter of the rectangular garden will be $=2(length+width)=2(x+y)$

Therefore, half of the perimeter of the garden will be $=x+y$

According to the question, the half of the perimeter is $36m$

Therefore,

$x+y=36\,\,\,\,\,\,\,\,\,\,……(i)$

Again,

The length of the garden is $4m$ more than its width.

$\therefore x=y+4\,\,\,\,\,\,\,\,\,\,……(ii)$

From Equation $(i)$

$x+y=36$

$\Rightarrow y=36-x$

Similarly, from Equation $(ii)$

$x=y+4$

$\Rightarrow y=x-4$

Plotting Table 1 and 2 we get,

The two lines intersect at the point $(20, 16)$.

Therefore, the length of the rectangular garden is $20m$ and width is $16m$.

## Question 6: Given the linear equation $2x+3y–8=0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

### i. Intersecting lines

Solution:

Given linear equation is $2x+3y-8=0$

Here, $a_1=2,\,b_1=3$ and $c_1=-8$

For intersecting lines, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

$\therefore \frac{2}{a_2}\ne \frac{3}{b_2}$

$a_2=5$ and $b_2=2$ satisfies the above condition. Therefore, the required equation in two variables is $5x+2y+2=0$.

### ii. Parallel lines

Solution:

Given linear equation is $2x+3y-8=0$

Here, $a_1=2,\,b_1=3$ and $c_1=-8$

For parallel lines, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\therefore \frac{2}{a_2}=\frac{3}{b_2}\ne \frac{-8}{c_2}$

$a_2=4,\,b_2=6$ and $c_2=5$ satisfies the above condition.

Therefore, the required equation in two variables is $4x+6y+5=0$.

### iii. Coincident lines

Solution:

Given linear equation is $2x+3y-8=0$

Here, $a_1=2,\,b_1=3$ and $c_1=-8$

For coincident lines, $\frac{a_1}{a_2}=\frac{b_1}{b_2}= \frac{c_1}{c_2}$

$\therefore \frac{2}{a_2}=\frac{3}{b_2}= \frac{-8}{c_2}$

$a_2=8,\,b_2=12$ and $c_2=-32$ satisfies the above condition.

Therefore, the required equation in two variables is $8x+12y-32=0$.

## Question 7: Draw the graphs of the equations $x–y+1=0$ and $3x+2y–12=0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:

Given the pair of linear equations are,

$x–y+1=0\,\,\,\,\,\,\,\,\,\,……(i)$

$3x+2y–12=0\,\,\,\,\,\,\,\,\,\,(ii)$

From Equation $(i)$

$x-y+1=0$

$\Rightarrow y=x+1$

Similarly, from Equation $(ii)$

$3x+2y-12=0$

$\Rightarrow 2y=12-3x$

$\Rightarrow y=\frac{12-3x}{2}$

Plotting Tables 1 and 2 we get,

The coordinates of the vertices of the triangle are $(2,3),\,(4,0)$ and $(-1,0)$.

The two lines cuts the x-axis at $(-1,0)$ and $(4,0)$ respectively.

### NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2 PDF Download

The PDF of NCERT solutions for Class 10 Maths Chapter 3, Exercise 3.2 is provided below.

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### List of Exercises in Pair of Linear Equations in Two Variables, Chapter 3, Class 10 Maths

• Exercise 3.1 (NCERT Solutions for Ex. 3.1 Class 10 Maths, Chapter 3)
• Exercise 3.2 (NCERT Solutions for Ex. 3.2 Class 10 Maths, Chapter 3)
• Exercise 3.3 (NCERT Solutions for Ex. 3.3 Class 10 Maths, Chapter 3)
• Exercise 3.4 (NCERT Solutions for Ex. 3.4 Class 10 Maths, Chapter 3)
• Exercise 3.5 (NCERT Solutions for Ex. 3.5 Class 10 Maths, Chapter 3)
• Exercise 3.6 (NCERT Solutions for Ex. 3.6 Class 10 Maths, Chapter 3)
• Exercise 3.7 (NCERT Solutions for Ex. 3.7 Class 10 Maths, Chapter 3)

I hope the NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear.

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### What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 3?

The main topics covered in the NCERT Solutions for Class 10 Maths, Chapter 3 are: Pair of Linear Equations in Two Variables, Graphical Method of Solution of a Pair of Linear Equations, Algebraic Methods of Solving a Pair of Linear Equations, Substitution Method, Elimination Method, Cross-Multiplication Method and Equations Reducible to a Pair of Linear Equations in Two Variables.

### How many exercises are there in the NCERT Solutions for Class 10 Maths Chapter 3?

There are 7 exercises in the NCERT Solutions for Class 10 Maths Chapter 3.

### Is the NCERT solutions for Class 10 Maths Chapter 3 important for exam point of view?

Yes, the NCERT solutions for Class 10 Maths Chapter 3 are very much important for exam point of view as it covers very basic concepts on Pair of Linear Equations in Two Variables.

### Is Exercise 3.2 is of NCERT solutions for Class 10 Maths important for examination point of view?

Yes, Exercise 3.2 of NCERT solutions for Class 10 Maths is very important for examination point of view as covers the topic Graphical Method of Solution of a Pair of Linear Equations.