NCERT Solutions Class 10 - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1

Chapter 3 of the Class 10 NCERT Mathematics textbook deals with the Pairs of Linear Equations in Two Variables. Exercise 3.1 focuses on solving and understanding the basic concepts of such equations. This chapter is essential for mastering techniques used to find solutions to systems of equations and is foundational for the more advanced algebraic concepts.

What is a Pair of Linear Equations in Two Variables?

A pair of linear equations in the two variables consists of the two linear equations involving the two variables typically x and y. These equations can be written in the general form:

a1x+b1y=c1

𝑎2𝑥+𝑏2𝑦=𝑐2

where a1,b1,c1,a2,b2 and c2 are constants. The solution to this pair of equations is the set of values (x,y) that simultaneously satisfy both equations. Graphically, these equations represent lines on the coordinate plane and their point of intersection is the solution to the system.

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

Solution:

Let's take,

Number of girls = x

Number of boys = y

According to the given conditions,

x + y = 10            -(1)

x – y = 4             -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

x + y = 10, So, we can use the following table to draw the graph:

x	y
0	10
10	0

For equation (2)

x - y = 4, So, we can use the following table to draw the graph:

x	y
0	-4
4	0

The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (7, 3).

Hence, the number of girls are 7 and number of boys are 3 in a class.

(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's take,

Cost for one pencil = x

Cost for one pencil = y

According to the given conditions,

5x + 7y = 50             -(1)

7x + 5y = 46             -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

5x + 7y = 50, So, we can use the following table to draw the graph:

x	y
3	5
10	0

For equation (2)

7x + 5y = 46, So, we can use the following table to draw the graph:

x	y
3	5
8	-2

The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (3, 5).

Hence, the cost of a pencil is ₹ 3 and cost of a pen is ₹ 5.

Question 2. On comparing the ratios \frac{a1}{a2}, \frac{b1}{b2} , and \frac{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

Solution:

In the given equations, 

a1 = 5

a2 = 7

b1 = -4

b2 = 6

c1 = 8

c2 = -9

Now, here

a1/a2 = 5/7

b1/b2 = -4/6 = -2/3

c1/c2 = 8/-9

As, here

\frac{a1}{a2} ≠ \frac{b1}{b2}  

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

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

Solution:

In the given equations,

a1 = 9

a2 = 18

b1 = 3

b2 = 6

c1 = 12

c2 = 24

Now, here

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = 12/24 = 1/2

As, here

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}  

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

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

Solution:

In the given equations,

a1 = 6

a2 = 2

b1 = -3

b2 = -1

c1 = 10

c2 = 9

Now, here

a1/a2 = 6/2 = 3

b1/b2 = -3/-1 = 3

c1/c2 = 10/9

As, here

\frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}  

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

Question 3. On comparing the ratios \frac{a1}{a2}, \frac{b1}{b2} , and \frac{c1}{c2} , find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5; 2x – 3y = 7 

Solution:

In the given equations,

a1 = 3

a2 = 2

b1 = 2

b2 = -3

c1 = -5

c2 = -7

Now, here

a1/a2 = 3/2

b1/b2 = 2/-3

c1/c2 = -5/-7

As, here

\frac{a1}{a2} ≠ \frac{b1}{b2}

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

Pair of linear equations are CONSISTENT.

(ii) 2x – 3y = 8; 4x – 6y = 9

Solution:

In the given equations,

a1 = 2

a2 = 4

b1 = -3

b2 = -6

c1 = -8

c2 = -9

Now, here

a1/a2 = 2/4 = 1/2

b1/b2 = -3/-6 = 1/2

c1/c2 = -8/-9 = 8/9

As, here

\frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}  

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

Pair of linear equations are INCONSISTENT.

(iii) \frac{3}{2}x + \frac{5}{3}y  = 7; 9x – 10y = 14

Solution:

In the given equations,

a1 = 3/2 

a2 = 9

b1 = 5/3 

b2 = -10

c1 = -7

c2 = -14

Now, here

\frac{a1}{a2} = \frac{\frac{3}{2}}{9} = \frac{1}{6}

\frac{b1}{b2} = \frac{\frac{5}{3}}{-10} = \frac{-1}{6}

c1/c2 = -7/-14 = 1/2

As, here

\frac{a1}{a2} ≠ \frac{b1}{b2}

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

Pair of linear equations are CONSISTENT.

(iv) 5x – 3y = 11; – 10x + 6y = –22

Solution:

In the given equations,

a1 = 5

a2 = -10

b1 = -3

b2 = 6

c1 = -11

c2 = 22

Now, here

a1/a2 = 5/-10 = -1/2

b1/b2 = -3/6 = -1/2

c1/c2 = -11/22 = -1/2

As, here

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}  

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

(v) \frac{4}{3}x  + 2y = 8; 2x + 3y = 12

Solution:

In the given equations,

a1 = 4/3 

a2 = 2

b1 = 2

b2 = 3

c1 = -8

c2 = -12

Now, here

\frac{a1}{a2} = \frac{\frac{4}{3} }{2} = \frac{2}{3}

b1/b2 = 2/3

c1/c2 = -8/-12 = 2/3

As, here

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}  

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

Question 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

Solution:

In the given equations,

a1 = 1

a2 = 2

b1 = 1

b2 = 2

c1 = -5

c2 = -10

Now, here

a1/a2 = 1/2

b1/b2 = 1/2

c1/c2 = -5/-10 = 1/2

As, here

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}  

Hence, the given pairs of equations have infinite many solutions and the lines are coincident.

Pair of linear equations are CONSISTENT.

x + y = 5             -(1)

2x + 2y = 10             -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

x + y = 5, So, we can use the following table to draw the graph:

x	y
0	5
5	0

For equation (2)

2x + 2y = 10, So, we can use the following table to draw the graph:

x	y
0	5
5	0

The graph will be as follows for Equation (1) and (2):

(ii) x – y = 8, 3x – 3y = 16

Solution:

In the given equations,

a1 = 1

a2 = 3

b1 = -1

b2 = -3

c1 = -8

c2 = -16

Now, here

a1/a2 = 1/3

b1/b2 = 1/3

c1/c2 = -8/-16 = 1/2

As, here

\frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}  

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

Pair of linear equations are INCONSISTENT.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

Solution:

In the given equations,

a1 = 2

a2 = 4

b1 = 1

b2 = -2

c1 = -6

c2 = -4

Now, here

a1/a2 = 2/4 = 1/2

b1/b2 = 1/-2

c1/c2 = -6/-4 = 3/2

As, here

\frac{a1}{a2} ≠ \frac{b1}{b2}

Hence, the given pairs of equations have a unique solution and the lines intersect each other at exactly one point.

Pair of linear equations are CONSISTENT.

2x + y – 6 = 0             -(1)

4x – 2y – 4 = 0              -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

2x + y – 6 = 0, So, we can use the following table to draw the graph:

x	y
0	6
3	0

For equation (2)

4x – 2y – 4 = 0, So, we can use the following table to draw the graph:

x	y
0	-2
1	0

The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (2, 2).

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:

In the given equations,

a1 = 2

a2 = 4

b1 = -2

b2 = -4

c1 = -2

c2 = -5

Now, here

a1/a2 = 2/4 = 1/2

b1/b2 = -2/-4 = 1/2

c1/c2 = -2/-5 = 2/5

As, here

\frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}  

Hence, the given pairs of equations have no solution and the lines are parallel and never intersect each other.

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's take,

length = x

breadth = y

Half the perimeter of a rectangular garden = \frac{2(x+y)}{2}   = x + y

According to the given conditions,

x = y + 4             -(1)

x + y = 36              -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

x = y + 4, So, we can use the following table to draw the graph:

x	y
0	-4
4	0

For equation (2)

x + y = 36, So, we can use the following table to draw the graph:

x	y
0	36
36	0

The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (20, 16).

Hence, length is 20 m and breadth is 16 m of rectangle.

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:

Linear equation in two variables such that pair so formed is intersecting lines, so it should satisfy the given conditions

\frac{a1}{a2} ≠ \frac{b1}{b2}

By rearranging, we get

\frac{a1}{b1} ≠ \frac{a2}{b2}

Hence, the required equation should not be in ratio of 2/3 

Hence, another equation can be 2x – 9y + 9 = 0

where the ratio is 2/-9 

and, \frac{2}{3} ≠ \frac{2}{-9}

(ii) parallel lines

Solution:

Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions

\frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}

By rearranging, we get

\frac{a1}{b1} = \frac{a2}{b2}

\frac{b1}{c1} ≠ \frac{b2}{c2}

Hence, the required equation a2/b2 should be in ratio of 2/3 and b2/c2 should not be equal to 3/-8 

Hence, another equation can be 4x + 6y + 9 = 0

where the ratio a2/b2 is 2/3

and, \frac{b2}{c2} ≠ \frac{3}{-8}

(iii) coincident lines

Solution:

Linear equation in two variables such that pair so formed is parallel lines, so it should satisfy the given conditions

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}

By rearranging, we get

\frac{a1}{b1} = \frac{a2}{b2}

\frac{b1}{c1} = \frac{b2}{c2}

Hence, the required equation a2/b2 should be in ratio of 2/3 and b2/c2 should be equal to 3/-8 

Hence, another equation can be 4x + 6y -16 = 0

where the ratio a2/b2 is 2/3 

and, b2/c2 = 3/-8

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:

x – y + 1 = 0              -(1)

3x + 2y – 12 = 0              -(2)

So, to construct a graph, we need to find at least two solutions of the given equation.  

For equation (1)

x – y + 1 = 0, So, we can use the following table to draw the graph:

x	y
0	-1
1	0

For equation (2)

3x + 2y – 12 = 0, So, we can use the following table to draw the graph:

x	y
0	6
4	0

The graph will be as follows for Equation (1) and (2):

Now, from the graph, we can conclude the given lines intersect each other at point (2, 3), and x-axis at (−1, 0) and (4, 0). 

Hence, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

Summary

Exercise 3.1 of Chapter 3 (Pair of Linear Equations in Two Variables) in NCERT Class 10 Mathematics introduces the concept of linear equations in two variables and their graphical representation. It covers topics such as identifying linear equations, determining whether a given pair of equations forms a consistent or inconsistent system, and graphing linear equations. Students learn to plot linear equations on a coordinate plane, interpret the nature of their solutions based on the graphs, and understand the geometric significance of different types of solutions (one solution, no solution, or infinitely many solutions).

Pair of Linear Equations in Two Variables

What is a linear equation in two variables?

A linear equation in two variables is an equation that can be written in the form ax + by + c = 0, where a, b, and c are constants, and a and b are not both zero. It represents a straight line when graphed on a coordinate plane.

How do you determine if two linear equations represent the same line?

Two linear equations represent the same line if they are equivalent after simplification. This means they have the same slope and y-intercept, or in other words, they are scalar multiples of each other.

What does it mean for a system of linear equations to be consistent or inconsistent?

A system of linear equations is consistent if it has at least one solution (one or infinitely many). It is inconsistent if it has no solution. Graphically, consistent equations intersect at least once, while inconsistent equations are parallel and never intersect.

What is the difference between dependent and independent linear equations?

Independent equations represent different lines and have a unique solution (one intersection point). Dependent equations represent the same line and have infinitely many solutions (the lines overlap completely).