NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables

NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables

NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables: National Council of Educational Research and Training Class 10 Maths Chapter 3 Solutions – Pair Of Linear Equations In Two Variables. NCERT Solutions Class 10 Maths Chapter 3 PDF Download.

NCERT Solutions Class 10 Maths Chapter 3: Overview

Board NCERT
Class 10
Subject Maths
Chapter 3
Chapter Name Pair Of Linear Equations In Two Variables
Topic Exercise Solutions

 

NCERT Solutions Class 10 Maths

Chapter 3Pair Of Linear Equations In Two Variables

 

Exercise 3.1

(1) Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically

Solution:

Let the present age of Aftab be x years.

—- let the correct age if his daughter be y year.

Seven years ago

Age of Aftab = (x – 7) years

Age of his daughter = (y – 7) years

According to the question

(x – 7) = 7 (y – 7

=> x – 7 = 7y – 49

=> x – 7y = -42 —– (1)

Also, Three years later

Age of Aftab = (x + 3) year

Age of Aftab’s daughter = (y + 3) years

According to the question

(x + 3) = 3 (y + 3)

=> x + 3 = 3y + 9

=> x – 3y = 6 —– (2)

For equation (1)

x – 7y = – 42

Let, x = 0

0 – 7y = -42

=> 7y = 42

=> y = 42/7

=> y = 6

The solution is (0, 6)

For equation (2)

x – 3y = 6

Let, x = 0

-3y = 6

y = 6/3

y = -2

The solution is (0, -2)  

X 0 7
Y 6 7

 

Let, x = 7

x – (7×7) = -42

=> x – 49 = -42

=> x = 7

The solution is (7, 7)

X 0 6
Y -2 0

 

Let y = 0

x = 6

The solution is (6, 0)

 

(2) The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.

Solution:

Let, the price of a bat be RS x and the price of a ball be RS y

For 1st case

3x + 6y = 3900

=> 3 (x + 2y) = 3900

=> (x + 2y) = 1300 —— (1)

Now, x + 3y = 130 —– (2 )

For equation (1)

x + 2y = 1300

Let x = 0

0 + 2y = 1300

=> 2y = 1300

=> y = 1300/2

=> y = 650

x 0 100
y 650 600

  

Let, x = 100

100 + 2y = 1300

=> 2y = 1200

=> y = 1200/2

=> y = 600

The solutions are (0, 650), (100, 600) for equation (2)

x + 3y = 130

Let, x = 0

0 + 3y = 1300

=> y = 1300/3

=> y = 433.33

X 0 100
y 433 400

Let, x = 100

100 + 3y = 1300

=> 3y = 1300 – 100

=> 3y = 1200

=> y = 1200/3

=> y = 400

The solution are = (0, 433.33), (100, 400)

 

(3) The cost of 2 kg of apples and 1kg of grapes on a day was found to be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.

Solution:

Let, the cost of one kg apple be RS x act the cost of one kg grapes be Rs y it’s given that ‘

2x + y = 160 —— (1)

Also, 4x + 2y = 300

=> 2 (2x + y) = 300

=> x + y = 300/2

=> 2x + y = 150 —– (2)

For equation (1)

2x + y = 160

Let, x = 50

2 (50) + y = 160

=> 100 + y = 160

=> y = 160 – 100

=> y = 60

x 50 80
y 60 0

 

Let, y = 0

2x + 0 = 160 

=> x = 160/2

=> x = 80

The solution of equation (1) is (50, 60), (80, 0)

For equation (2)

2x + y = 150

Let, x = 50

(2×50) + y = 150

=> 100 + y = 150

=> y = 150 – 100

=> y = 50

x 50 60
y 50 30

 

Let, x = 60

(2×60) + y = 150

= 120 + y = 150

= y = 150 – 120

= y = 30

The solution for equation (2) are (50, 50), (60, 30|)

 

Exercise – 3.2

 

(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:

Total number of student is 10

Let, the numbers of boys be x and the number of girls be y

Given that,

x + y = 10 —— (1)

And also y = x + 4

=> y – x = 4 —- (2)

For equation (1)

x + y = 10

Let x = 0

 0 + y = 10

= y = 10

x 0 10
y 10 0

 

Let, y = 0

x + 0 = 10

=> x = 10

The solution for equation (1) are (0, 10), (10, 0)

For equation (2)

y – x = 4

Let, x = 0

y – 0 = 4

=> y = 4

X 0 4
y 4 8

Let, x = 4

y – 4 = 4

=> y = 4+4

=> y = 8

The solutions in equation (2) are (0, 4), (4, 8)

The two equations are hold at (3, 7 )

The solution is (3, 7)

The number of girls in the quiz competion are 3 and The number of girls in the quiz competion are 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 price of one pencil be RS x and the price one pen be Rs y

Given that, 5x + 7y = 50 —— (1)

and also, 7x + 5y = 46 —– (2)

For equation (1)

5x + 7y = 50

Let, x = 10

5 × (10) + 7y = 50

=> 50 + 7y = 50

=> 7y = 0

y = 0

x 10 3
y 0 5

Let, x = 3

(5×3) + 7y = 50

=> 15 + 7y = 50

=> 7y = 50 – 15

=> 7y = 35

=> y = 35/7

=> y = 5

The solution for equation (1) are (10,0) (3, 5)

For equation (2)

7x + 5y = 46

Let, x = 3

(7×3) + 5y = 46

=> 21 + 5y = 46

=> 5y = 46 – 21

=> 5y = 25

=> y = 25/5

= y = 5

x 3 -2
y 5 12

 

x = (-2)

7 (-2) + 5y = 46

=> -14 + 5y = 46

=> 5y = 46 + 14

=> 5y = 60

=> y = 60/5

=> y = 12

The solution for equation (2) is (3, 5), (-2, 12)

The two line intersect at (3, 5)

The cost of one pencil is RS 3 and the cost of one pen is RS 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

Solution:

5x – 4y + 8 = 0 ——- (1)

7x + 6y – 9 = 0 ——- (2)

For equation (1)

5x – 4y + 8 = 0

Comparing with a1 x + b1 y + c1 = 0

a1 = 5, b1 = -4, c1 = 8

For equation (2)

7x + 6y – 9 = 0

Comparing with a2 x + b2 y + c2 = 0

a2 = 7, b2 = 6, c2 = -9

a1/a2 = 5/7

b1/b2 = -4/6

= -2/3

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

Since, a1/a2 ≠ b1/b2

We have unique solution for two equations.

The lines that represent the linear equation intersects at a point

 

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

18x + 6y + 24 = 0

Solution:

9x + 3y + 12 = 0 —– (1)

18x + 6y + 24 = 0 ——- (2)

For equation (1)

9x + 3y + 12 = 0

Comparing with

a1 x + b1 y + c1 = 0

a1 = 9, b1 = 3, c1 = 12

For equation (2

18x + 6y + 24 = 0

Comparing with

a2 x + b2 y + c2 = 0

a2 = 18, b2 = 6, c2 = 24

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

So, we have infinite solutions

The lines that respect the linear equation are coincident.

 

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solution:

6x – 3y + 10 = 0 —– (1)

2x – y + 9 = 0 —— (2)

For equation (1)

6x – 3y + 10 = 0

Comparing with,

a1 x + b1 y + c1 = 0

a1 = 6, b1 = -3, c1 = 10

For equation (2)

2x – y + 9 = 0

Comparing with,

a2 x + b2 x + c2 = 0

a2 = 2, b2 = -1, c2 = 9

a1/a2 = 6/2

= 3/1

b1/b2 = -3/-1

c1/c2 = 10/9

So,We have no solution

The lines represent the linear equations are parallel.

 

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

(i) 3x + 2y = 5; 2x – 3y = 7

Solution:

3x + 2y – 5 = 0 —- (1)

2x – 2y – 7 = 0 —– (2)

For equation (1)

3x + 2y – 5 = 0

comparing with

a1 x + b1 y + c1 = 0

a1 = 3, b1 = 3, c1 = 5

For equation (2)

2x – 3y – 7 = 0

Comparing with

a2 x + b2 y + c2 = 0

a2 = 2, b2 = -3, c2 = -7

a1/a2 = 3/2

b1/b2 = 3/-3

= -1/1

c1/c2 = -5/-7 = 5/7

Since, a1/a2 ≠ b1/b2

So, we have unique solution

our system is consistent

 

(ii) 2x – 3y = 8; 4x – 6y = 9

Solution:

2x – 3y – 8 = 0 —— (1) 

4x – 6y – 9 = 0 —— (2)

For equation (1)

2x – 3y – 8 = 0

Comparing with

a1 x + b1 y + c1

a1 = 2, b1 = -3, c1 = -8

For equation (2)

4x – 6y – 9 = 0

Comparing with

a2 x + b2 y + c1 = 0

a2 = 4, b2 = 6, c2 = -9

a1/a2 = 2/4

= 1/2 

b1/b2 = -3/-6

= 1/2 

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

Since, a1/a2 = b1/b2 ≠ c1/c2

We have no solution

Our system is inconsistent

 

(iii) 3/2 x 5/3 y = 7; 9x – 10y = 14

Solution:

3/2 x + 5/3 y = 7

=> 3/2 x + 5/3 y – 7 = 0 ——- (1)

9x – 10y = 14

=> 9x – 10y – 14 = 0 —– (2)

For equation (1)

3/2 x + 5/3 y – 7 = 0

Comparing with

a1 x + b1 y + c2 = 0

a1 = 3/2, b1 = 5/3, c1 = -7

For equation (2)

9x – 10y – 14 = 0

Comparing with

a2 x + b2 y + c2 = 0

a2 = 9, b2  -10, c2 = -14

a1/a2 = 3/2×4

= 1/6

b1/b2 = 5/3 × 102

= -1/6

c1/c2 = -7/-14

= 1/2 

Since a1/a2 ≠ b1/b2

So, we have unique solution

Our system is consistent

 

(iv) 5x – 3y = 11; – 10x + 6y = –22

Solution:

5x – 3y = 11

=> 5x – 3y – 11 = 0 —– (1)

-10x + 6y = -22

=> – 10x + 6y + 22 = 0 —– (2)

For equation (1)

5x – 3y – 11 = 0

Comparing with

a1 x + b1 y + c1 = 0

a1 = 5, b1 = -3, c1 = -11

For equation (2)

-10x + 6y + 22 = 0

Comparing with –

a2 x + b2 y + c2 = 0

a2 = -10, b2 = 6, c2 = 22

a1/a2 = 5/-10

b1/b2 = -3/6

= -1/1

c1/c2 = -11/22

= -1/2

Sine, a1/a2 = b1/b2 = c1/c2

So, we have infinitely many solutions

Our system is consistent

 

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

Solution:

4/3 x + 2y = 8

=> 4/3 + 2y – 8 = 0 —— (1)

2x + 3y = 12

=> 2x + 3y – 12 = 0 —— (2)

For equation (1)

4/3 x + 2y – 8 = 0

Comparing with

a1 x + b1 y + c1 = 0

a1 = 4/3, b1 = 2, c2 = -8

For equation (2)

2x + 3y – 12 = 0

Comparing with

a2 x + b2 y + c2 = 0 

a2 = 2, b2 = 3, c2 = -12

a1/a2 = 4/3×2

= 2/3

b1/b2 = 2/3

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

Since, a1/a2 = b1/b2 = c1/c2

So, we have infinitely many solutions our system is 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

Solution:

x + y = 5

=> x + y – 5 = 0 —— (1)

2x + 2y = 10

=> 2x + 2 – 10 = 0  

For equation (1)

x + y – 5 =  0

Comparing with a1 x + b1 y + c1

a1 = 1, b1 = 1, c1 = 5

For equation (2)

2x + 2y – 10 = 0

Comparing with a2 x + b2 y + c2 = 0

a2 = 2, b2 = a2, c2 = -10

a1/a2 = 1/2 

b1/b2 = 1/2 

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

Since, a1/a2 = b1/b2 = c1/c2

So, we have infinitely many solution

The system is consistent

Now solving the equation graphically

For equation (1)

x + y = 5

Let, x = 0

0 + y = 5

y = 5

x 0 5
y 5 0

 

Let, y = 0

x + 0 = 5 

= x = 5

The solutions for equation are (0, 5), (5, 0)

For equation (2)

2x + 2y = 10

Let, x = 0

0 + 2y = 10

=> 2y = 10

=> y = 10/2 = 5

X 0 5
y 5 0

 

Let, y = 0

2x = 10

=> x = 5

The solutions are (0, 5), (5, 0)

 

(ii) x – y = 8, 3x – 3y = 16

Solution:

x – y = 8 —-

=> x – y – 8 = 0 —— 0

3x – 3y = 16

=> 3x – 3y – 16 = 0 —– (2)

For equation (1)

x – y – 8 = 0

Comparing with a1 x + b1 y + c1 = 0

a1 = 1, b1 = -1, c1 = –

For equation = (2)

3x – 3y – 16 = 0

Comparing both a2 x + b2 y + c2 = -8

For equation (2)

3x – 3y – 16 = 0

Comparing with a2 x =3, b2 = -3, c2 = -16

a2/a2 = 1/3

b1/b2 = 1/-3 = 1/3

c1/c2 = -8/-16 = ½ 

Since, a1/a2 = b1/b2 ≠ c1/c

So, we have no solution

The system is inconsistent

 

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

Solution:

2x + y – 6 = 0 —— (1)

4x – 2y – 4 = 0 —— (2)

For equation (1)

2x + y – 6 = 0

Comparing with

a1 x + b1 y + c1 = 0

a1 = 2, b1 = 1, c1 = -6

For equation (2)

4x – 2y – 4 = 0

Comparing with

a2 x + b2 y + c2 = 0

a2 = 4, b2 = -2,  c2 = -4

a1/a2 = 2/4, = 1/2 

b1/b2 = 1/-2 

= 1/-2

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

Since, a1/a2 ≠ b1/b2

So we have unique solution

The system in consistent

Now solving the equation graphically for equation (1)

2x + y – 6 = 0

Let, x = 0

0 + y – 6 = 0

y = 6

x 0 3
y 6 0

 

Let, y = 0

2x – 6 = 0

= 2x = 6

The solutions for equation (1) is (0, 6), (3, 0)

For equation (2)

4x – 2y – 4 = 0 

Let, x = 0

-2y – 4 = 0

-2y = 4

y = -2

x 0 1
y -2 0

 

Let, y = 0

4x – 4 = 0

4x = 4

x = 1

The solutions equation (2) are (0, -2), (1, 0)

 

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:

2x – 2y – 4 = 0 —— (1)

4x – 4y – 5 = 0 ——- (2)

For equation (1)

2x – 2y – 4 = 0

Comparing with –

a1 x + b1 y + c1 = 0

a1 = 2, b1 = -2, c1 = -4

For equation (2)

4x – 4y – 5 = 0

Comparing with

a2 x + b2 y + c2 = 0

a2 = 4, b2 = -4, c2 = -5

a1/a2 = 2/4 = 1/2 

b1/b2 = -2/-4 = 1/2 

c1/c2 = -4/-5 = 4/5

Since, a1/a2 = b1/b2 ≠ c1/c2

So, we have no solutions

The system 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 = x meters and the breadth = y meters

Give that,

½ x 2 (x + y) = 36

=> x + y = 36 —– (1)

and also,

x = y + 4 

=> x – y =  4

 

For Equation –

x + y = 36

Let x = 16

6 + y = 16

y = 36 – 16

y = 20

x 16 20
y 20 16

 

Let, y = 16

x + 16 = 36

=> x = 36 – 16

=> x = 20

Solutions for equation (1) are (16, 20), (20, 16)

For equation (2)

x – y = 4

Let, x = 4

4 – y = 4

=> -y = 4 – 4

=> y = 0

Let, y = 4

x – 4 = 4 

=> x = 4 + 4

=> x = 8 

Solutions for equation (2) are (4, 6), (8, 4) Putting both equation on the graph

Both lines intersect each other at (20, 16)

The solutions of two equation is (20, 16)

The length of the rectangular garden is 20m and breadth is 16m

 

(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:

Given that

2x + 3y – 8 = 0 —— (1)

Comparing with equation a, x + b1 y + c1 = 0

a1 = 2, b1 = 3, c1 = -8

(i) Let the equation for intersecting lines

a2 x + b2 y + c2 = 0

So, a1/a2 ≠ b1/b2

As, a1 ≠ b1

a2 = 1, b2 = 1 and c2 = 1 also the equation for intersecting lines is x + y + 1 = 0

 

(ii) Let, the equation for parallel lines a3 x + b3 y + c3 = 0

So, a1/a3 = b1/b3 ≠ c1/c3

Grph Page Photo 20

a1/a3 = b1/b3

=> 2/a3 = 3/b3

=> a3/b3 = 2/3 

a3 = 2

b3 = 3

c1/c3 ≠ a1/a3

c1/c3 ≠ 2/2

c1/c3 ≠ 1

c3 = 1, or c1 = -8

The equation for parallel lines is

2x + 3y + 1 = 0

 

(iii) Let the equation of coincident lines is –

a4 x + b4 y + c4 = 0

So, a1/a4 = b1/b4 = c1/c4

2/a4 = 3/b4 = -8/c4

a4 = 4, b4 = 6, c4 = -16

The equation for co-incident line is – 4x + 6y – 16 = 0

 

(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 equations

x – y = -1 —– (1)

3x + 2y = 12 —– (2)

For equation (-1)

x – b = – 1

Let, x = 0

0 – b = 1

=> -b = -1

=> b = 1 

x 0 4
y 6 0

 

Let, y = 0

x – 0 = -1

x = -1

The solutions for equation (1) are (0, 1), (-1, 0)

For equation (2)

3x + 2y = 12

Let, x = 0

0 + 2y = 12

=> 2y = 12

=> y = 12/2

=> y = 6

x 0 4
y 6 0

 

Let, y = 0 

3x + 0 = 12

=> 3x = 12 

=> x = 12/3

=> x = 4

The solutions for equations (2) are (0, 6), (4, 0, 6)

Plotting with equation on graph

The vertices required to form triangle are (2, 3), (-1, 0) and (4, 0)

 

Exercise – 3.3

 

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

(i) x + y = 14

x – y = 4

Solution:

x + y = 14 —— (1)

x – y = 4 —– (2)

From equation (1) —

x + y = 14

= x = 14 – y —- (3)

Substituting value of x in equation (2)

x – y = 4

(14 – y) – y = 4

=> 14 – y – y = 4 

=> 14 – 2y = 4

=> 2y = 14 – 4

=> 2y = 10

=> y = 10/2

Putting y = 5 in equation (2)

x – 5 = 4 

=> x = 4+5

=> x = 9

x = 9 and y = 5 are the solution for the equation.   

 

(ii)  s – t = 3

s/3 + t/2 = 6

Solution:

s-t = 3  —— (1)

s/3 + t/2 = 6 —— (2)

s – t = 3

s = 3 + t

Substituting value of s in equation (2)

3+t/3 + t/2 = 6

 

Graph page photo 23

 

2 (3+t) +3t/6 = 6

=> 6 + 2t + 3t = 36

=> 5t = 36-6

=> 5t = 30

=> t = 30/5 = 6 = t = 6

Putting t = 6 in equation (1)

s – 6 = 3

=> s = 3+6

=> s = 9

s = 9, t = 6  

 

(iii) 3x – y = 3

9x – 3y = 9

Solution:

3x – y = 3 —— (1) 

9x – 3y = 9

=> 3 (3x – y) = 9

=> 3x – y = 9/3

=> 3x – y = 3 —– (2)

Equation (1) = equation (2)

For all true value of x there are infinitely many solutions.

 

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

Solution:

0.2x + 0.3y = 1.3 —– (1)

0.4x + 0.5y = 2.3 —– (2)

0.2x + 0.3y = 1.3

=> 0.2x = 1.3 – 0.3y

=> x = 1.3-0.3y/0.2

=> x = (13-3y) × 10/×10

=> x = 13-3y/2 —— (3)

Substituting value of x in equation (2)

0.4 × 13-3y/2 + 0.5y = 2.3 

Multiplying both sides by 10

10 x 0.4 × 13-3y/2 + 0.5 × 10y = 2.3 × 10

=> 4 × 13-3y/2 + 5y = 23

=> 26 – 6y + 5y = 23

=> -y = 23 – 26

=> -y = -3

= y = 3

Putting y = 3 in equation (3)

x = 13 – (3×3)/2

13 – 9/2 = 4/2 = 2

x = 2 —- y = 3 

 

(v) √2x + √3 y = 0

√3x – √8y = 0

Solution:

√2 x + √3 y = 0 —— (1)

√3 x – √8 y = 0 —— (2)

√2 x + √3 y = 0

√2 x = – √3 y

=> x = -√3/√2 y ——- (3)

Substituting value of x in equation (2)

-√3/√2 × √3y -√8y = 0

=> – 3/√2 y = √8 y

=> -3y = √8 × √2 y = 0

=> -3y – √16 y = 0

=> -3y – 4y = 0

=> -7y = 0

=> y = 0

Putting y = 0 in equation (3)

x = -√3/√2 × 0

x = 0 and y = 0

 

(vi) 3x/2 – 5y/3 = -2

x/3 + y/2 = 13/6

Solution:

3x/2 – 5y/3 = -2

=> 9x -10y/6 = -2 

=> 9x – 10y = -12 —— (1)

 

x/3 + y/2 = 13/6

2x+3y/6 = 13/6

=> 2x + 3y = 13 —– (2)

 

9x – 10y = -12

=> 9x = -12 + 10y

=> x = -12+10y/9 —— (3)   

Substituting value of x in equation (2)

2 × -12+10y/9 + 3y = 13

=> -24+20y+27y/9 = 13

=> -24+20y+27y = 13×9

=> 47y = 117 + 24

=> 47y = 141/47

=> y = 3

Putting y = 3 in equation (3)

x = -12+(10×3)/9

=> -12+30/9

=> 18/9 = 2

x = 2, y = 3 is the solution

 

(2) Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11 —– (1)

2x – 4y = -24 —– (2)

2x + 3y = 11

=> 2x = 11-3y

=> x = 11-3y/2 —– (3)

Substituting value of x in equation (2)

2 × 11-3y/2 – 4y = -24

=> 11-3y – 4y = -24

=> -7y = -24-11

=> -7y = -35

=> 7y = 35

=> y = 35/7

Putting y = 5 in equation (3)

 x = 11 – (3×5)/2

=> 11-15/2 = 4/2 = -2

x = -2 and y = 5 is the solution of the & equation

Now, y = mx + 3

Putting the value of x = -2 and y = 5

5 = m (-2) + 3

=> 5 – 3 = m (-1)

=> 2 = -2 m

=> m = 2/-2 = -1

The value of m is-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 26 and one number is three times the other. Find them.

Solution:

Let, the larger number = x and the smaller number = y

Give that,

x – y = 26 —– (1)

and also given that x = 3y —- (2)

Substituting value of x in equation (i)

3y – y = 26

=> 2y = 26

=> y = 26/2 = 13

Putting y = 13 in equation (2)

x = 3 × 13 = 39

x = 39 and y = 1

The number are 39 and 13

 

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution: –

Let, the bigger angle = x and the smaller angle = y

Given that,

x – y = 18 —— (1)

Now both the angles are supplementary

x + y = 180 —– (2)

From equation (i)

x – y = 18

=> x = 18+y —– (3)

Substituting x in equation (2)

x + y = 180

=> (18 + y) + y = 180

=> 18 + 2y = 180

=> 2y = 180 – 18

=> 2y = 162

=> y = 162/2

=> y = 81

Putting y = 81 in (3)

x = 18 + 81

= 94 

x = 99 and y = 81

The bigger angle x = 99°

The smaller angle y = 81°

 

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Solution:

Let, the price of one bat = Rs x

and the price of one ball = Rs y

Given that,

7x + 6y = 3800 —– (1)

Also,

3x + 5y = 1750 —– (2)

From equation (1)

7x + 6y = 3800

=> 7x = 3800 – 6y

=> 7x = 3800-6y/7 —- (3)

Substituting x in equation (2)

3 (3800-6y)/7) + 5y = 1750

=> 11400-18y+35y/7 = 1750

=> 11400 – 18y + 35y = 7×1750

=> 17y = 12250 – 11400

=> 17y = 850

=> y = 850/17

=> y = 50

Putting y = 5 in equation (3)

x = 3800 – 6×50/7

=> 3800 – 300/7

=> 3500/7 = 500

x = 500 and y = 50

The cost of one bat = Rs 500 and cost of one ball Rs 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 10 km, the charge paid is 105 and for a journey of 15 km, the charge paid is 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution: 

Let, the fixed change = RS x and change per km = Rs y

Given, that total change for 10km is Rs 105

x + 10y = 105 —- (1)

And also total change for 15km is Rs 155

x + 15y = 155 —- (2)

From equation (1)

x + 10y = 105

x = 105 – 10y — (2)

Substituting x in equation (2)

(105 – 10y) + 15y = 155

=> 105 – 10y + 15y = 155

=> 5y = 155 – 105

=> 5y = 50

=> y = 50/5 = 10

Putting y = 10 in equation

x = 105 – 100

x = 5

x = 5 and y = 10

Fixed changes is Rs 5 and charge per km Rs 10

Now, the cost for a person to travel 25 km is

= Rs 5 + (25×10)

= RS 5 + 250

= RS 255

 

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator.

If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:

Let, the numerator be x

And the denominator be y

Given that,

x+2/y+2 = 9/11

=> 11 (x+2) = 9 (y+2)

=> 11x + 22 = 9y + 18

=> 11x – 9y = 18 – 22

=> 11x – 9y = -4 —– (1)

Also,

x+3/y+3 = 5/6

=> 6 (x + 3) = 5 (y + 3)

=> 6x + 18 = 5y + 15 

=> 6x – 5y = 15 – 18

=> 6x – 5y = -3 —– (2)

From equation (1)

11x – 9y = -4

= 11x = -4 + 9y

= x = -4+9y/11 —– (3)

Substituting x in equation (2)

6 × -4+9y/11 – 5y = -3

=> -24+54y-55y/11 = -3

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

=> -y = -33+24

=> -y = -9

=> y = 9

Putting y = 9 in equation (3)

x = -4+81/11

x = 77/11 = 7

x = 7, y = 9

The original fraction = x/y = 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?

Solution:

Let, current age of Jacob’s son = y years

Five years later

Jacob’s age = x + 5 years 

Jacob son’s age = y + 5 years

Given,

(x + 5) = 3 (y + 5)

=> x + 5 = 3y + 15

=> x – 3y = 10 —- (1)

Five years age

Jacob’s age = x – 5 years

Jacob’s son’s age = y – 5 years

Given,

(x – 5) = 7(y – 5)

=> x – 5 = 7y – 35

=> x – 7y = -35+5

=> x – 7y = -30 —- (2)

From equation (1)

x – 3y = 10

=> x = 10 + 3y —– (3)

Substituting x in the equation (2)

10 + 3y – 7y = -30

=> -4y = -30 – 10

=> -4y = -40

=> 4y = 40

=> y = 10

Putting y = 10 in equation (3)

x = 10 + (3×10) 

=> 10 + 30 = 40

x = 40 and y = 1

Present age of Jacob is 40 years and age of his son is 10 years.

 

Exercise – 3.4

 

(1) Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

Solution:

x + y = 5 —– (1)

2x – 3y = 4 —— (2)

Multiplying equation (1) by (2)

2 (x + y) = 2×5

= 2x + 2y = 10 —— (3)

Solving (3) and (2) by elimination doing (3) – (2)

(3) = 2x + 2y = 10
(2) = 2x – 3y = 4
(-)      (+)    (-)
__________________
(Substraction) 5y = 6

y = 6/5 

Putting y = 6/5 is equation (1)

x + 6/5 = 5

=> x = 5 – 6/5

=> x = 25-6/5 = 19/5

x = 19/5 and y = 6/5

 

(ii) 3x + 4y = 10 and 2x – 2y = 2

Solution:

3x + 4y = 10 —- (1)

2x – 2y = 2 —– (2)

Multiplying equation (2) by (2)

2 (2x – 2y) = 2×2

=> 4x – 4y = 4 —– (3)

Solving (3) and (2) by elimination

(3) = 3x + 4y = 10
(4) = 4x – 4y = 4
(+)     (-)
_________________
(Addition) 7x = 14

=> x = 14/7 = 2

Putting x = 2 in equation (2)

4 – 2y = 2

=> -2y = -2

=> y = 1

x = 2, y = 1    

Also, from equation (2)

2x – 2y = 2

2 (x – y) = 2

x – y = 1

x = 1 + y —– (4)

Substituting x in equation (1)

3 (1+y) + 4y = 10

=> 3 + 3 + 4 = 10

=> 7y = 10-3

=> y = 7/7 = 1

y = 1

Putting y = 1 in equation (4)

x = 1 + 1 = 2

x = 2 and y = 1

 

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Solution:

3x – 5y – 4 = 0

=> 3x – 5y = 4 —– (1)

9x = 2y + 7

=> 9x – 2y = 7 —– (2)

Multiplying equation (1) by (3)

3 (3x – 5y) = 4×3

=> 9x – 15y = 12 —- (3)

Solving (3) and (2) by elimination method

(3) => 9x – 15y = 12
(2) => 9x – 2y = 7
(-)   (+)      (-)
___________________
(Substraction) – 13y = 5 

= y = -5/13

Putting y = -5/13 in equation (2)

9x – 2 (-5/13) = 7

=> 9x + 10/13 = 7

=> 9x = 7 – 10/13

=> 9x = 91-10/13

=> 9x = 81/13

=> x = 9/13

x = 9/13 and y = -5/13

From equation (1)

3x – 5y = 4

=> 3x = 4 + 5y

=> x = 4+5y/3

Substituting x in equation (2)

3x × 4+5y/3 -2y = 7

=> 12 + 15y – 2y = 7

=> 13y = 7-12

=> 13y = -5

=> y = – 5/13

Putting y = – 5/13 in equation (2)

9x – 2 × (-5/13) = 7

=> 9x + 10/13 = 7

=> 9x = 7 – 10/13

=> 9x = 91-10/13

=> 9x = 81/13

=> x = 81/13 × 1/9

=> x = 9/13

x = 9/13 and y = – 5/13

 

(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

Solution:

x/2 + 2y/3 = -1

=> 3x + 4y/6 = 1

=> 3x + 4y = -6 —– (1)

x – y/3 = 3

=> 3x – y/3 = 3

=> 3x – y = 9 —– (2)

Solving (1) and (2) by elimination

(2) = 3x + 4y = -6
(2) = 3x – y = 9
(-)   (+) 
 (-)
_________________
5y = -15

=> y = -15/5 = -3

Putting y = -3 in equation (2)

3x – (-3) = 9

=> 3x + 3 = 9

=> 3x = 9 – 3

=> 3x = 6

=> x = 6/3 = 2 

x = 2, y = -3

From equation (2)

3x – y = 9

=> 3x = 9 + y

=> x = 9+y/3

Substituting value of x in equation (1)

3 × 9+y/3 + 4y = -6

=> 9 + y + 4y = -6

=> 5y = – 6 -9

=> 5y = -15

=> y = – 15/5

= – 3

x = 2 and y = 3 is the solution. 

 

(2) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Solution:

Let, the numerator be x and denominator be y

Give that,

x + 1/y – 1 = 1

=> x + 1 = y – 1

=> x – y = -1 -1

=> x – y = -2 —– (1)

Also given that

x/y+1 = 1/2 

=> 2x = y + 1

=> 2x – y = 1

Solving (1) and (2) by elimination method

(1) = x – y = -2
(2) = 2x – y = -1
(-)   (+)
________________
– x = -3

=> -x = -3

=> x = 3

Putting x = 3 in equation (1)

3 – y = -2

– y = -2 -3

=> -y = -5

=> y = 5

x = 3, y = 5

The fraction is 3/5

 

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

Let, the current age of Nuri = x years 

And current age of Sonu = y year 

Five, years ago

Age of Nuri = (x – 5) year

Age of Sonu = (y – 5) years

Given that, (x – 5) = 3 (y – 5)

=> x – 5 = 3y – 15

=> x – 3y = -15 +5

=> x – 3y = -10 —– (i)

Ten years later,

Age of Nuri = (x + 10) years

Age of Sonu = (y + 10) years

Also, gives that

x + 10 = 2 (y + 10)

=> x + 10 = 2y + 20

=> x – 2y = 20 – 10

=> x – 2y = 10 —- (2)

Solving (1) and (2) by elimination method

(1) = x – 3y = -10
(2) = x – 2y = 10
(-) (+)   (-)
_________________
-y = -20

= y = 20

Putting y = 20 in equation (2)

x – (2×20) = 10

9x – 40 = 10

=> x = 10 + 40

=> x = 50

The present age of Nuri is 50 years and present age of Sonu is 20 years.

 

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution

Let, the digit in units place is x

The number in tens place is y

∴ The number in the form yx

The number is (10 × y) + x = 10y + x

Given that x + y = 9

=> y + x = 9 ——- (1)

Also = y + x = 9 —- (1)

9 (10y + x) = 2 × (10x + y)

=> 90y + 9x = 20x + 2y

=> 90y – 2y = +9x -20x = 0

=> 88y – 11x = 0

=> 11 (8y – x) = 0

=> 8y – x = 0 —– (2)

Solving equation (2) and (1 by elimination method

(2) = 8y – x = 0
(1) = y + x = 9
_______________
(Addition) 9y = 9

y = 1

Putting y:1 in equation (1)

1 + x = 9

x = 9 – 1

= 8

The number is {(1× 10)} = 10 + 8 = 18

 

(iv) Meena went to a bank to withdraw 2000. She asked the cashier to give her 50 and 100 notes only. Meena got 25 notes in all. Find how many notes of 50 and 100 she received.

Solution:

Let, there are x number of Rs 50 notes an y number of Rs 100 notes

Given that

x + y = 25 — (1)

And also given that

50x + 100y = 2000

=> 50 (x + 2y) = 2000

=> x + 2y = 2000/50

x + 2y = 40 —– (2)

Solving equation (1) and (2) by elimination method

(1) x + y = 25
(2) x + 2y = 40
(-)   (-)     (-)
______________
-y = -25

= y = 25

Putting y = 25 in equation (1)

x + 15 = 25

x = 25 – 15

= x = 10

There are 10 of Rs 50 notes and 15 of Rs 100 notes.

 

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Updated: December 16, 2021 — 11:22 am

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