# NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.6) ## Ex 3.6 Question 1.

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

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

(iii) 4/x + 3y = 14

3/x -4y = 23

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Solution:

(i) 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let 1/x = p and 1/y = q ,

p/2 + q/3 = 2

⇒ 3p+2q-12 = 0………….(1)

p/3 + q/2 = 13/6

⇒ 2p+3q-13 = 0………….(2)

Using cross-multiplication method,

p/(-26-(-36) ) = q/(-24-(-39)) = 1/(9-4)

p/10 = q/15 = 1/5

p/10 = 1/5 and q/15 = 1/5

So, p = 2 and q = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Let 1/√x = p and 1/√y = q

2p + 3q = 2 ………..(i)

4p – 9q = -1 ………(ii)

Multiplying equation (i) by 3,

6p + 9q = 6 …….…..(iii)

10p = 5

p = 1/2…….…(iv)

Now by putting the value of ‘p’ in equation (i),

2×1/2 + 3q = 2

3q = 1

q = 1/3

p =1/√x

½ = 1/√x

x = 4

q = 1/√y

1/3 = 1/√y

y = 9

x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

4m + 3y – 14 = 0  …..…..(i)

3m – 4y – 23 = 0  ………….(ii)

By cross-multiplication,

p/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-p/125 = y/50 = -1/ 25

-p/125 = -1/25 and y/50 = -1/25

p = 5 and y = -2

p = 1/x = 5

x = 1/5

y = -2

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Let 1/(x-1) = p and 1/(y-2) = q

5p + q = 2 ….………(i)

6p – 3q = 1 ………..….(ii)

Multiplying equation (i) by 3,

15p + 3q = 6 ………….(iii)

21p = 7

p = 1/3

Putting this value in equation (i),

5×1/3 + q = 2

q = 2- 5/3 = 1/3

p = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

⇒ x = 4

q = 1/(y-2)

⇒ 1/3 = 1/(y-2)

⇒ y = 5

x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

(7x-2y)/ xy = 5

7/y – 2/x = 5…………..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15…………(ii)

Let 1/x = p , 1/y = q

-2p + 7q – 5 = 0  ……..(iii)

7p + 8q – 15 = 0 ……(iv)

By cross-multiplication method,

p/(-105-(-40)) = q/(-35-30) = 1/(-16-49)

p/(-65) = q/(-65) = 1/(-65)

p/-65 = 1/-65

p = 1

q/(-65) = 1/(-65)

q = 1

p = 1 and n = 1

p = 1/x = 1       q = 1/x = 1

x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = p and 1/y = q

⇒ 6q +3p = 6

⇒3p + 6q -6 = 0……….(i)

2x + 4y = 5xy

⇒ 2/y + 4/x = 5

⇒ 2q +4p = 5

⇒ 4p+2q-5 = 0………..(ii)

3p + 6q – 6 = 0

4p + 2q – 5 = 0

By cross-multiplication method,

p/(-30 –(-12)) = q/(-24-(-15)) = 1/(6-24)

p/-18 = q/-9 = 1/-18

p/-18 = 1/-18

p = 1

q/-9 = 1/-18

q = 1/2

p = 1 and q = 1/2

p= 1/x = 1 and q = 1/y = 1/2

x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Let 1/x+y = p and 1/x-y = q

10m + 2n = 4      ⇒ 10m + 2n – 4 = 0      …..…..(i)

15m – 5n = -2    ⇒   15m – 5n + 2 = 0    …………..(ii)

Using cross-multiplication method,

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

x+y = 5 …………………(iii)

q = 1/(x-y) = 1

x-y = 1…………………(iv)

2x = 6  ⇒ x = 3 …….(v)

Putting the value of x = 3 in equation (iii),

y = 2

x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Let 1/(3x+y) = p and 1/(3x-y) = q

p+ q= 3/4 …….…… (i)

p/2 – q/2 = -1/8

p – q = -1/4  …………..…(ii)

2p = 3/4 – 1/4

2p = 1/2

Putting in (ii),

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)

6x = 6

x = 1 …………….(v)

Putting in (iii),

3(1) + y = 4

y = 1

x = 1 and y = 1

## Ex 3.6 Question 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.

(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.

(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 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.

Solutions:

(i) Let the speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Speed of Ritu while rowing upstream = (x-y) km/h

Speed of Ritu while rowing downstream = (x + y) km /h

2(x+y) = 20

⇒ x + y = 10…….(i)

2(x-y) = 4

⇒ x – y = 2………(ii)

2x = 12

x = 6

Putting the value of x in equation (i)

y = 4

∴Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let the number of days taken by a woman and a man to finish the work be x and y respectively.

Work done by women in one day = 1/x

Work done by man in one day = 1/y

4(2/x + 5/y) = 1

⇒(2/x + 5/y) = 1/4

3(3/x + 6/y) = 1

⇒(3/x + 6/y) = 1/3

Now, put 1/x=p and 1/y=q,

2p + 5q = 1/4 ⇒ 8p + 20q = 1…………………(i)

3m + 6n =1/3 ⇒ 9p + 18q = 1………………….(ii)

Using cross multiplication method,

p/(20-18) = q/(9-8) = 1/ (180-144)

p/2 = q/1 = 1/36

p/2 = 1/36

p = 1/18

p = 1/x = 1/18

x = 18

q = 1/y = 1/36

y = 36

∴Number of days taken by women to finish the work = 18

Number of days taken by men to finish the work = 36.

(iii) Let the speed of the train = x km/h

Speed of the bus = y km/h

60/x + 240/y = 4 …………………(i)

100/x + 200/y = 25/6 …………….(ii)

Put 1/x=p and 1/y=q,

60p + 240q = 4……………………..(iii)

100p + 200q = 25/6

600p + 1200q = 25 ………………….(iv)

Multiply equation (iii) by 10,

600p + 2400q = 40 …………(v)

Subtracting equation (iv) from (v),

1200q = 15

q = 15/1200 = 1/80

Putting the value of n in equation (iii)

60p + 3 = 4

p = 1/60

p = 1/x = 1/60

x = 60

And y = 1/n

y = 80

∴Speed of the train = 60 km/h

Speed of the bus = 80 km/h