# Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.6 pdf || UP Board

# __Pair of Linear Equations in Two
Variables__

__Exercise – 3.6__

__Pair of Linear Equations in Two Variables__

__Exercise – 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)

Adding
equations (6) and (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.

Download Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.6 pdf

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