Class 10th:Lesson 3 Pair of Linear Equations in Two Variables Maths

Class 10th: Lesson 3 Pair of Linear Equations in Two Variables Math

      Ex 3.1

       
  Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2





 

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

Answer

(i) x + y = 5; 2x + 2y = 10
a₁/a₂ = 1/2
b₁/b₂ = 1/2 and
c₁/c₂ = 5/10 = 1/2
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5 - y

x	4	3	2
y	1	2	3

And, 2x + 2y = 10
x = 10-2y/2

x	4	3	2
y	1	2	3

Graphical representation

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) x – y = 8, 3x – 3y = 16
a₁/a₂ = 1/3
b₁/b₂ = -1/-3 = 1/3 and
c₁/c₂ = 8/16 = 1/2
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -1/2 and
c₁/c₂ = -6/-4 = 3/2
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y - 6 = 0
y = 6 - 2x

x	0	1	2
y	6	4	2

And, 4x - 2y -4 = 0
y = 4x - 4/2

x	1	2	3
y	0	2	4

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -2/-4 = 1/2 and
c₁/c₂ = 2/5
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, these linear equations are parallel to each other and thus, have no possible solution. Hence, 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.

Answer

Let length of rectangle = x m
Width of the rectangle = y m
According to the question,
y - x = 4 ... (i)
y + x = 36 ... (ii)
y - x = 4
y = x + 4

x	0	8	12
y	4	12	16

y + x = 36

x	0	36	16
y	36	0	20

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

6. Given the linear equation 2x + 3y - 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines

(ii) parallel lines
(iii) coincident lines

Answer

(i) Intersecting lines:
For this condition,
a₁/a₂ ≠ b₁/b₂
The second line such that it is intersecting the given line is
2x + 4y - 6 = 0 as
a₁/a₂ = 2/2 = 1
b₁/b₂ = 3/4 and
a₁/a₂ ≠ b₁/b₂

(ii) Parallel lines

For this condition,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hence, the second line can be
4x + 6y - 8 = 0 as
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = -8/-8 = 1
and a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(iii) Coincident lines
For coincident lines,
a₁/a₂ = b₁/b_2 = c₁/c₂
Hence, the second line can be
6x + 9y - 24 = 0 as
a₁/a₂ = 2/6 = 1/3
b₁/b₂ = 3/9 = 1/3 and
c₁/c₂ = -8/-24 = 1/3
and a₁/a₂ = b₁/b_2 = c₁/c₂


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.

Answer

x - y + 1 = 0
x = y - 1

x	0	1	2
y	1	2	3

3x + 2y - 12 = 0

x = 12 - 2y/3

x	4	2	0
y	0	3	6

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( - 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( - 1, 0), and (4, 0).

Page No: 53
   

Exercise 3.3

1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14 ; x – y = 4

(ii) s – t = 3 ; s/3 + t/2 = 6
(iii) 3x – y = 3 ; 9x – 3y = 9 

(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0 ; √3x - √8y = 0 

(vi) 3/2x - 5/3y = -2 ; x/3 + y/2 = 13/6

Answer

(i) x + y = 14 ... (i)

x – y = 4 ... (ii)
From equation (i), we get

x = 14 - y ... (iii)
Putting this value in equation (ii), we get

(14 - y) - y = 4

14 - 2y = 4

10 = 2y

y = 5 ... (iv)

Putting this in equation (iii), we get

x = 9

∴ x = 9 and y = 5

(ii) s – t = 3 ... (i)
s/3 + t/2 = 6 ... (ii)
From equation (i), we gets = t + 3
Putting this value in equation (ii), we get
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5 ... (iv)
Putting in equation (iii), we obtain
s = 9
∴ s = 9, t = 6

(iii) 3x - y = 3 ... (i)
9x - 3y = 9 ... (ii)
From equation (i), we get
y = 3x - 3 ... (iii)
Putting this value in equation (ii), we get
9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x - 3
Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3 ... (i) 
0.4x + 0.5y = 2.3 ... (ii)
0.2x + 0.3y = 1.3
Solving equation (i), we get
0.2x = 1.3 – 0.3y
Dividing by 0.2, we get
x = 1.3/0.2 - 0.3/0.2
x = 6.5 – 1.5 y …(iii)
Putting the value in equation (ii), we get
0.4x + 0.5y = 2.3
(6.5 – 1.5y) × 0.4x + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
-0.1y = 2.3 – 2.6
y = -0.3/-0.1
y = 3
Putting this value in equation (iii) we get
x = 6.5 – 1.5 y
x = 6.5 – 1.5(3)
x = 6.5 - 4.5
x = 2
∴ x = 2 and y = 3

(vi) 3/2x - 5/3y = -2 ... (i)

x/3 + y/2 = 13/6 ... (ii)

From equation (i), we get

9x - 10y = -12

x = -12 + 10y/9 ... (iii)

Putting this value in equation (ii), we get

2. Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which y =mx + 3.

Answer

2x + 3y = 11 ... (i)
Subtracting 3y both side we get
2x = 11 – 3y … (ii)
Putting this value in equation second we get
2x – 4y = – 24 … (iii)
11- 3y – 4y = - 24
7y = - 24 – 11
-7y = - 35
y = - 35/-7
y = 5
Putting this value in equation (iii) we get
2x = 11 – 3 × 5
2x = 11- 15
2x = - 4
Dividing by 2 we get
x = - 2
Putting the value of x and y
y = mx + 3.
5 = -2m +3
2m = 3 – 5
m = -2/2
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 26 and one number is three times the other. Find them.

Answer

Let larger number = x
Smaller number = y
The difference between two numbers is 26
x – y = 26
x = 26 + y
Given that one number is three times the other
So x = 3y
Putting the value of x we get
26y = 3y
-2y = - 2 6
y = 13
So value of x = 3y
Putting value of y, we get
x = 3 × 13 = 39
Hence the numbers are 13 and 39.

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

Answer

Let first angle = x
And second number = y
As both angles are supplementary so that sum will 180
x + y = 180
x = 180 - y ... (i)
Difference is 18 degree so that
x – y = 18
Putting the value of x we get
180 – y – y = 18
- 2y = -162
y = -162/-2
y = 81
Putting the value back in equation (i), we get
x = 180 – 81 = 99Hence, the angles are 99º and 81º.

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

Answer

Let cost of each bat = Rs x
Cost of each ball = Rs y

Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
7x + 6y = 3800
6y = 3800 – 7x
Dividing by 6, we get
y = (3800 – 7x)/6 … (i)

Given that she buys 3 bats and 5 balls for Rs 1750 later.
3x + 5y = 1750
Putting the value of y
3x + 5 ((3800 – 7x)/6) = 1750
Multiplying by 6, we get
18x + 19000 – 35x = 10500
-17x =10500 - 19000
-17x = -8500
x = - 8500/- 17
x = 500
Putting this value in equation (i) we get
y = ( 3800 – 7 × 500)/6
y = 300/6
y = 50

Hence cost of each bat = Rs 500 and cost of each balls = 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 Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Answer

Let the fixed charge for taxi = Rs x
And variable cost per km = Rs y
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10y = 105 … (i)
x = 105 – 10y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15y = 155
Putting the value of x we get
105 – 10y + 15y = 155
5y = 155 – 105
5y = 50
Dividing by 5, we get
y = 50/5 = 10
Putting this value in equation (i) we get
x = 105 – 10 × 10
x = 5
People have to pay for traveling a distance of 25 km
= x + 25y
= 5 + 25 × 10
= 5 + 250
=255

A person have to pay Rs 255 for 25 Km.

(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.

Answer

Let Numerator = x
Denominator = y
Fraction will = x/y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x + 2)/y+2 = 9/11

By Cross multiplication, we get
11x + 22 = 9y + 18
Subtracting 22 both side, we get
11x = 9y – 4
Dividing by 11, we get
x = 9y – 4/11 … (i)
Given that 3 is added to both the numerator and the denominator it becomes 5/6.
If, 3 is added to both the numerator and the denominator it becomes 5/6
(x+3)/y +3  = 5/6 … (ii)
By Cross multiplication, we get
6x + 18 = 5y + 15
Subtracting the value of x, we get
6(9y – 4 )/11 + 18 = 5y + 15
Subtract 18 both side we get
6(9y – 4 )/11 = 5y - 3
54 – 24 = 55y - 33
-y = -9
y = 9
Putting this value of y in equation (i), we get
x = 9y – 4
11 … (i)
x = (81 – 4)/77
x = 77/11
x = 7

Hence our fraction is 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?

Answer

Let present age of Jacob = x year
And present Age of his son is = y year
Five years hence,
Age of Jacob will = x + 5 year
Age of his son will = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5 = 3(y + 5)
Adding 5 both side, we get
x = 3y + 15 - 5
x = 3y + 10 … (i)
Five years ago,
Age of Jacob will = x - 5 year
Age of his son will = y - 5 year
Jacob’s age was seven times that of his son
x – 5 = 7(y -5)
Putting the value of x from equation (i) we get
3y + 10 – 5 = 7y – 35
3y + 5 = 7y – 35
3y – 7y = -35 – 5
-4y = - 40
y = - 40/- 4
y = 10 year
Putting the value of y in equation first we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

Page No: 56
  
 Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

   
   Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5



Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6



   vii) 10/x+y + 2/x-y = 4
15/x+y - 5/x-y = -2
Putting 1/x+y = p and 1/x-y = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
⇒ 15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/x-y = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3x-y = 3/4
1/2(3x-y) - 1/2(3x-y) = -1/8
Putting 1/3x+y = p and 1/3x-y = q in the given equations, we get
p + q = 3/4 ... (i)
p/2 - q/2 = -1/8
p - q = -1/4 ... (ii)
Adding (i) and (ii), we get
2p = 3/4 - 1/4
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 - q = -1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x+y = 1/4
3x + y = 4 ... (iii)
q = 1/3x-y = 1/2
3x - y = 2 ... (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Answer

Let the speed of Ritu in still water and the speed of stream be x km/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream = (x - y) km/h

Downstream = (x + y) km/h

According to question,

2(x + y) = 20

⇒ x + y = 10 ... (i)
2(x - y) = 4
⇒ x - y = 2 ... (ii)
Adding equation (i) and (ii), we get

Putting this equation in (i), we get

y = 4

Hence, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Answer

Let the number of days taken by a woman and a man be x and y respectively.
Therefore, work done by a woman in 1 day = 1/x

According to the question,

4(2/x + 5/y) = 1

2/x + 5/y = 1/4

3(3/x + 6/y) = 1

3/x + 6/y = 1/3

Putting 1/x = p and 1/y = q in these equations, we get

2p + 5q = 1/4

By cross multiplication, we get

p/-20-(-18) = q/-9-(-18) = 1/144-180

p/-2 = q/-1 = 1/-36

p/-2 = -1/36 and q/-1 = 1/-36

p = 1/18 and q = 1/36

p = 1/x = 1/18 and q = 1/y = 1/36

x = 18 and y = 36

Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer

Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,

60/u + 240/v = 4 ... (i)
100/u + 200/v = 25/6 ... (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 ... (iii)
100p + 200q = 25/6
600p + 1200q = 25 ... (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 .... (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 ... (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.