# Pair of Linear Equations in Two VariablesExercise – 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

Let, 1/x = a and 1/y = b …………(3)

From equation (1),

a/2 + b/3 = 2

3a + 2b = 12

From equation (2),

a/3 + b/2 = 13/6

2a + 3b = 13 ………….(5)

Substituting the value of a from Equation (4),

24 – 4b + 9b = 39

5b = 15

b = 3

Substituting the value of b in Eq. (5),

2a + 3×3 = 13

2a = 13 – 9

2a = 4

a = 2

From equation (3),

1/x = 2, 1/y = 3

x = ½, y = 1/3

Hence x = ½, y = 1/3.

Let, 1/√x = a and 1/√y = b …………(3)

From equation (1),

2a + 3b = 2

a = (2 – 3b)/2 ………………….(4)

From equation (2),

4a – 9b = -1 ………….(5)

Substituting the value of a from Equation (4),

4 – 6b - 9b = -1

-15b = -5

b = 1/3

Substituting the value of b in Eq. (5),

4a - 9×1/3 = -1

4a – 3 = -1

4a = 2

a = 1/2

From Eq. (3),

1/√x = 1/2, 1/√y = 1/3

√x = 2, √y = 3

x = 4, y = 9

Hence x = 4, y = 9.

Let, 1/x = a …………(3)

From Eq. (1),

4a + 3y = 14

From Eq. (2),

3a – 4y = 23 ………….(5)

Substituting the value of a from Equation (4),

42 – 9y – 16y = 92

-25y = 50

y = -2

Substituting the value of y in Eq. (5),

3a - 4×-2 = 23

3a = 23 - 8

3a = 15

a = 5

From Eq. (3),

1/x = 5

x = 1/5

Hence x = 1/5, y = -2.

Let, 1/(x-1) = a and 1/(y-2) = b …………(3)

From Eq. (1),

5a + b = 2

b = 2 – 5a………………….(4)

From Eq. (2),

6a – 3b = 1 ………….(5)

Substituting the value of b from Equation (4),

6a – 3(2 – 5a) = 1

6a – 6 + 15a = 1

21a = 7

a = 1/3

Substituting the value of a in Eq. (5),

6×1/3 – 3b = 1

2 = 3b + 1

1 = 3b

b = 1/3

From Eq. (3),

1/(x-1) = 1/3, 1/(y-2) = 1/3

x - 1 = 3, y - 2 = 3

Hence x = 4, y = 5.

Let, 1/x = a and 1/y = b …………(3)

From Eq. (1),

7b – 2a = 5

7b = 5 + 2a

From Eq. (2),

8b + 7a = 15 ………….(5)

Substituting the value of b from Equation (4),

40 + 16a + 49a = 105

65a = 65

a = 1

Substituting the value of a in Eq. (5),

8b + 7×1 = 15

8b = 15 – 7

8b = 8

b = 1

From Eq. (3),

1/x = 1, 1/y = 1

x = 1, y = 1

Hence x = 1, y = 1.

Let, 1/x = a and 1/y = b …………(3)

From Eq. (1),

6b + 3a = 6

6b = 6 - 3a

From Eq. (2),

2b + 4a = 5 ………….(5)

Substituting the value of b from Equation (4),

12 - 6a + 24a = 30

18a = 18

a = 1

Substituting the value of a in Eq. (5),

2b + 4×1 = 5

2b = 5 – 4

b = 1/2

From Eq. (3),

1/x = 1, 1/y = 1/2

x = 1, y = 2

Hence x = 1, y = 2.

Let, 1/(x+y) = a and 1/(x-y) = b …………(3)

From Eq. (1),

10a + 2b = 4

2b = 4 - 10a

From Eq. (2),

15a – 5b = -2 ………….(5)

Substituting the value of b from Equation (4),

30a – 20 + 50a = -4

80a = 16

a = 1/5

Substituting the value of a in Eq. (5),

15×1/5 – 5b = -2

-5b = -3 –2

-5b = -5

b = 1

From Eq. (3),

1/(x+y) = 1/5, 1/(x-y) = 1

x + y = 5…………..(6), x - y = 1……………(7)

On adding Eq. (6) and (7),

2x = 5 + 1

2x = 6

x = 3

From Eq. (6),

3 + y = 5

y = 5 – 3

y = 2

Hence x = 3, y = 2.

Let, 1/(3x+y) = a and 1/(3x-y) = b …………(3)

From Eq. (1),

a + b = 3/4

b = 3/4 - a

From Eq. (2),

a/2 – b/2 = -1/8 ………….(5)

Substituting the value of b from Equation (4),

4a - 3 + 4a = -1

8a = 2

a = 1/4

Substituting the value of a in equation (4),

From Eq. (3),

1/(3x+y) = 1/4, 1/(3x-y) = 1/2

3x+y = 4………..(6), 3x-y = 2………………(7)

6x = 4 + 2

6x = 6

x = 1

From Eq. (6),

3×1 + y = 4

3 + y = 4

y = 1

Hence x = 1, 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.

Solution:

Let Ritu's speed of swimming in still water = x km/h

and speed of stream = y km/h

So, downstream speed = x + y km/h Distance = 20 km Time = 2 hours

According to Question,

speed = distance / time

x + y = 20/2

x + y = 10

y = 10 – x ………….(1)

Speed ​​upstream = x – y km/h Distance = 4 km Time = 2 hours

x – y = 4/2

x – y = 2

Substituting the value of y from Equation (1),

x – (10 – x) = 2

2x = 12

x = 6

Substituting the value of x in equation (1),

y = 10 – 6 = 4

Hence, the speed of Ritu swimming in still water is 6 km/h and the speed of the stream 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.

Solution:

Let the days taken by a woman to complete the work = x

and days taken by one man to complete the work = y

So, work done by one man in one day = 1/x

Work done by a woman in one day = 1/y

according to the first condition,

according to the second condition,

Let, 1/x = a and 1/y = b …….(3)

From Eq. (1),

8a + 20b = 1

a = (1 – 20b)/8  ……………..(4)

From Eq. (2),

9a + 18b = 1 ………..(5)

Substituting the value of a from Equation (4),

9 – 180b + 144b = 8

-36b = -1

b = 1/36

Substituting the value of b in Eq. (5),

9a + 18(1/36) = 1

9a + ½ = 1

18a + 1 = 2

a = 1/18

From equation (3),

1/x = 1/18, 1/y = 1/36

Hence, it will take 18 days for a woman to complete the work and 36 days for a man to complete the work.

(iii) Roohi travels 300 km to her home party by train and partly by bus. She takes 4 hours if she travels 60 km by train and the 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.

Solution:

Let, speed of train = x km/h

and speed of bus = y km/h

If she travels 60 km by train and the rest by bus, she takes 4 hours.

speed = distance / time

time = distance/speed

According to Question,

If she travels 100 km by train and the rest by bus, she takes 10 minutes.

According to Question,

Let, 1/x = a and 1/y = b ……………(3)

From equation (1),

60a + 240b = 4

a = (4 – 240b)/60    ………………(4)

From equation (2),

100a + 200b = 25/6

Substituting the value of a from Equation (4),

40 – 2400b + 1200b = 25

-1200b = -15

b = 1/80

Substituting the value of b in Eq. (4),

From equation (3),

1/x = 1/60, 1/y = 1/80

x = 60, y = 80

Hence, the speed of the train is 60 km/h and speed of bus is 80 km/h.

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