NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.2

NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.2 is crucial to understanding the graphical approach to solving linear equations in two variables. This exercise gives a precise explanation of how to solve these equations through graphs. It is an important skill for Class 10 students. The explanation of the exercise provided here is in line with the new CBSE syllabus and assists students in understanding these concepts more deeply. Let's explore this basic topic to enhance your knowledge and problem-solving skills.

1.0Download NCERT Solutions for Class 10 Maths Chapter 3  Exercise 3.2 : Free PDF

2.0Introduction to Pair of Linear Equations in Two Variables:

A pair of linear equations in two variables consists of two such equations. The solution to this system of linear equations is where the two lines intersect, denoting the solutions of x and y that apply to both of the equations. Each of these equations, involving two variables (usually denoted as x and y), can be written in the form:

$a_{1}x+b_{1}x+c_1=0$

$a_{2}x+b_{2}x+c_2=0$

Here, a₁, b₁, and c₁ are the constants of equation one, whereas, a₂, b₂, and c₂ are the constants of equation two, where a₁, b₁, a₂, b₂, $\ne$ 0. The values of the variables x and y represent the solutions of these equations. 

3.0Exercise 3.2 Overview: Key Concepts 

Types of Pairs of Linear Equations Based on Their Graphs: 

• Consistent Pair (One solution): If the lines intersect at exactly one point, the system has one solution, and the pair of equations is called consistent. 
• Inconsistent Pair (No solution): If the lines are parallel and never intersect, the system has no solution, and the pair of equations is referred to as inconsistent.
• Dependent Pair (Infinitely many solutions): If the lines overlap, then the system possesses an infinite number of solutions. This type of equation pair is also termed a consistent equation.

Graphical Method of Solving Linear Equations in Two Variables:

The behaviour of the lines gives us insights into the solutions of the system of equations, that are $a_{1}x+b_{1}x+c_1=0$ and $a_{2}x+b_{2}x+c_2=0$.

1. Intersecting Lines (One or unique Solution): When the two lines cross at one point, there is only one solution to the system of equations. The point of intersection is the point with coordinates x and y that satisfy both equations. For a pair of linear equations in two variables to be intersecting at one point, it satisfies the following condition: 

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

2. Coincident Lines (Infinitely many solutions): Coincident lines are a pair of lines which coincide with each other at each point, meaning they both are the same lines. Coincident lines have infinitely many solutions to the system and can be represented by the following relation: 

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

3. Parallel Lines (No solution): A pair of lines never intersect if they are parallel to one another, which ultimately means that the system of equations doesn’t have any solution for the system of equations. These lines can be represented by the following relation: 

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

Intersecting Lines and Solutions of Linear Equations: 

The exercise is focused on finding the unknown variables of a pair of linear equations in two variables with graphical methods. When we graph two linear equations on a coordinate plane, their point of intersection is the solution to the system of equations. This point satisfies both equations at the same time, giving the values of the variables that solve the system. The graphing technique for solving linear equations in two variables helps students understand how different lines intersect and determine the range of possible solutions (unique, infinite, or none).

4.0NCERT Class 10 Maths Chapter 3 Exercise 3.2: Detailed Solutions

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, find the number of boys and girls who took part in the quiz.

(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.

Sol. (i) Let the number of boys be x and the number of girls be y.

According to the given conditions:

x + y = 10 and y = x + 4

We get the required pair of linear equations as:

x + y - 10 = 0, x - y + 4 = 0

Graphical Solution:

x + y - 10 = 0  ...(1)

x - y + 4 = 0  ...(2)

From the graph, we have: x = 3, y = 7 as the common solution of the two linear equations.

Hence, the number of boys = 3 and the number of girls = 7.

(ii) Let the cost of 1 pencil be ₹ x and the cost of 1 pen be ₹ y.

5x + 7y = 50

7x + 5y = 46

Graphical solution:

5x + 7y = 50

y = (50 - 5x) / 7

7x + 5y = 46

y = (46 - 7x) / 5

From the graph, we have x = 3, y = 5.

Hence, the cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.

2. On comparing the ratios a₁/a₂, b₁/b₂, and c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel, or coincident.

(i) 5x - 4y + 8 = 0 ; 7x + 6y - 9 = 0

(ii) 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0

(iii) 6x - 3y + 10 = 0 ; 2x - y + 9 = 0

Solutions:

(i) 5x - 4y + 8 = 0  ...(1)

7x + 6y - 9 = 0  ...(2)

 a₁/a₂ = 5/7

 b₁/b₂ = -4/6 = -2/3

Since a₁/a₂ ≠ b₁/b₂, the lines represented by equations (1) and (2) intersect at a point.

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

a₁/a₂ = 9/18 = 1/2

b₁/b₂ = 3/6 = 1/2

c₁/c₂ = 12/24 = 1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines represented by the equations are coincident.

(iii) 6x - 3y + 10 = 0

2x - y + 9 = 0

a₁/a₂ = 6/2 = 3/1

b₁/b₂ = -3/-1 = 3/1

c₁/c₂ = 10/9

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines represented by the equations are parallel.

3. On comparing the ratios a₁/a₂, b₁/b₂, and c₁/c₂, find out whether the following pairs of linear equations are consistent or inconsistent.

(i) 3x + 2y = 5 ; 2x - 3y = 7

(ii) 2x - 3y = 8 ; 4x - 6y = 9

(iii) (3/2)x + (5/3)y = 7 ; 9x - 10y = 14

(iv) 5x - 3y = 11 ; -10x + 6y = -22

(v) (4/3)x + 2y = 8 ; 2x + 3y = 12

Solutions:

(i) 3x + 2y = 5 ; 2x - 3y = 7

Rewrite the equations in the form ax + by + c = 0:

3x + 2y - 5 = 0

2x - 3y - 7 = 0

Calculate the ratios:

a₁/a₂ = 3/2

b₁/b₂ = 2/(-3) = -2/3

Since a₁/a₂ ≠ b₁/b₂, the equations have a unique solution.

Therefore, the equations are consistent.

(ii) 2x - 3y = 8 ; 4x - 6y = 9

Calculate the ratios:

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = -3/(-6) = 1/2

c₁/c₂ = 8/9

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the equations have no solution.

Therefore, the equations are inconsistent.

(iii) (3/2)x + (5/3)y = 7 ; 9x - 10y = 14

Calculate the ratios:

a₁/a₂ = (3/2)/9 = 1/6

b₁/b₂ = (5/3)/(-10) = -1/6

Since a₁/a₂ ≠ b₁/b₂, the equations have a unique solution.

Therefore, the equations are consistent.

(iv) 5x - 3y = 11 ; -10x + 6y = -22

Calculate the ratios:

a₁/a₂ = 5/(-10) = -1/2

b₁/b₂ = -3/6 = -1/2

c₁/c₂ = 11/(-22) = -1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations have infinitely many solutions.

Therefore, the equations are consistent.

(v) (4/3)x + 2y = 8 ; 2x + 3y = 12

Calculate the ratios:

a₁/a₂ = (4/3)/2 = 2/3

b₁/b₂ = 2/3

c₁/c₂ = 8/12 = 2/3

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations have infinitely many solutions.

Therefore, the equations are consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.

(i) x + y = 5, 2x + 2y = 10

(ii) x - y = 8, 3x - 3y = 16

(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0

(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

Solutions:

(i) x + y = 5, 2x + 2y = 10

Comparing with the general form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have:

a₁ = 1, b₁ = 1, c₁ = -5

a₂ = 2, b₂ = 2, c₂ = -10

Calculating the ratios:

a₁/a₂ = 1/2

b₁/b₂ = 1/2

c₁/c₂ = -5/-10 = 1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the pair of linear equations is consistent and has infinitely many solutions (coincident lines).

The equations x + y = 5 and 2x + 2y = 10 are essentially the same.

Therefore the points are (1,4) and (3,2).

(ii) x - y = 8, 3x - 3y = 16

Comparing with the general form:

a₁ = 1, b₁ = -1, c₁ = -8

a₂ = 3, b₂ = -3, c₂ = -16

Calculating the ratios:

a₁/a₂ = 1/3

b₁/b₂ = -1/-3 = 1/3

c₁/c₂ = -8/-16 = 1/2

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and have no solution. Therefore, the pair of linear equations is inconsistent.

(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0

Comparing with the general form:

a₁ = 2, b₁ = 1, c₁ = -6

a₂ = 4, b₂ = -2, c₂ = -4

Calculating the ratios:

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 1/-2 = -1/2

c₁/c₂ = -6/-4 = 3/2

Since a₁/a₂ ≠ b₁/b₂, the lines intersect at a unique point. Therefore, the pair of linear equations is consistent.

From the graph, the intersection point is (2, 2). Therefore, x = 2, y = 2.

(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0

Comparing with the general form:

a₁ = 2, b₁ = -2, c₁ = -2

a₂ = 4, b₂ = -4, c₂ = -5

Calculating the ratios:

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = -2/-4 = 1/2

c₁/c₂ = -2/-5 = 2/5

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel and have no solution. Therefore, the pair of linear equations is inconsistent.

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 rectangular garden be 'l' meters and the breadth be 'b' meters.

Given that the length is 4 m more than its width. Therefore, we can write:

l = b + 4  ...(1)

The perimeter of a rectangle is given by 2(l + b).

It is also given that half the perimeter of the garden is 36 m. Thus:

(1/2)  [2(l + b)] = 36

l + b = 36  ...(2)

Now, we can substitute the value of 'l' from equation (1) into equation (2):

(b + 4) + b = 36

2b + 4 = 36

2b = 36 - 4

2b = 32

b = 32 / 2

b = 16 m

Now, substitute the value of 'b' back into equation (1) to find 'l':

l = b + 4

l = 16 + 4

l = 20 m

Therefore, the length of the garden is 20 meters and the breadth of the garden is 16 meters.

Thus, the length of the garden = 20 m and breadth of the garden = 16 m.

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

(ii) Parallel lines

(iii) Coincident lines

Solution:

(i) Intersecting lines

Given equation: 2x + 3y - 8 = 0

New equation: 3x + 2y + 4 = 0

Here, a1/a2 ≠ b1/b2 (2/3 ≠ 3/2).

Hence, the graph of the two equations will be two intersecting lines.

(ii) Parallel lines

Given equation: 2x + 3y - 8 = 0

New equation: 4x + 6y - 10 = 0

Here, a1/a2 = b1/b2 ≠ c1/c2 (2/4 = 3/6 ≠ -8/-10 or 1/2 = 1/2 ≠ 4/5).

Hence, the graph of the two equations will be two parallel lines.

(iii) Coincident lines

Given equation: 2x + 3y - 8 = 0

New equation: 4x + 6y - 16 = 0

Here, a1/a2 = b1/b2 = c1/c2 (2/4 = 3/6 = -8/-16 or 1/2 = 1/2 = 1/2).

Hence, the graph of the two equations will be two coincident lines.

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:

Equation 1: x - y + 1 = 0  =>  y = x + 1

Equation 2: 3x + 2y - 12 = 0  =>  2y = 12 - 3x  =>  y = (12 - 3x) / 2

The vertices of the triangle are A(2, 3), B(-1, 0), and C(4, 0).

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.2

• By working through Exercise 3.2 solutions, students enhance their analytical thinking and problem-solving abilities.
• Since CBSE exams often include questions directly from NCERT, practicing these solutions improves exam readiness.
• Exercise 3.2 focuses on solving pairs of linear equations using the graphical method.
• NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2, provides a strong conceptual base, improves problem-solving skills, and prepares students for exams.