# NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.6

NCERT Solutions for Class 10 Maths Exercise 3.6 Chapter 3 Linear Equations In Two Variables contains detailed study material related to the topics provided in the textbook. NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Variables is mentioned in the NCERT Solutions along with the answers to the exercises provided in the textbook. NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.6 contains the solutions to the problems provided in the chapter.

The stepwise solutions to all the Maths problems asked in the textbook are given in NCERT Solutions Class 10 Maths. The students should practise these exercises to analyze their shortcomings.

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## Access Other Exercise Solutions of Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

Exercise 3.1 Solutionsâ€“ 3 Questions
Exercise 3.2 Solutionsâ€“ 7 Questions
Exercise 3.3 Solutionsâ€“ 3 Questions
Exercise 3.4 Solutionsâ€“ 2 Questions
Exercise 3.5 Solutionsâ€“ 4 Questions
Exercise 3.7 Solutionsâ€“ 8 Questions

### Access Answers to NCERT Class 10 Maths Chapter 3- 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:

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

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

Solution:Â

Let us assume 1/x = m and 1/y = n Â , then the equation will change as follows.

m/2 + n/3 = 2

â‡’ 3m+2n-12 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

m/3 + n/2 = 13/6

â‡’ 2m+3n-13 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/âˆšx + 3/âˆšy = 2

4/âˆšx + 9/âˆšy = -1

Solution:

Substituting 1/âˆšx = m and 1/âˆšy = n in the given equations, we get

2m + 3n = 2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

4m â€“ 9n = -1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m =Â 1/2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦(iv)

Now by putting the value of â€˜mâ€™ in equation (i), we get

2Ã—1/2 + 3n = 2

3n = 1

n =Â 1/3

m =1/âˆšx

Â½ = 1/âˆšx

x = 4

n = 1/âˆšy

1/3 = 1/âˆšy

y = 9

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Solution:

PuttingÂ in the given equation we get,

So, 4m + 3y = 14Â Â Â Â  => 4m + 3y â€“ 14 = 0Â  â€¦â€¦â€¦â€¦â€¦..â€¦..(1)

3m â€“ 4y = 23Â Â Â Â  => 3m â€“ 4y â€“ 23 = 0Â  â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

By cross-multiplication, we get,

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

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

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

m = 5 and b = -2

m =Â 1/xÂ = 5

So , x = 1/5

y = -2

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

6/(x-1) â€“ 3/(y-2) = 1

Solution:

Substituting 1/(x-1) = m and 1/(y-2) = n Â in the given equations, we get,

5m + n = 2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

6m â€“ 3n = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)

Adding (ii) and (iii) we get

21m = 7

m =Â 1/3

Putting this value in equation (i), we get

5Ã—1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

â‡’ 1/3 = 1/(x-1)

â‡’ x = 4

n = 1/(y-2)

â‡’ 1/3 = 1/(y-2)

â‡’ y = 5

Hence, x = 4 and y = 5

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

(8x + 7y)/xy = 15

Solution:

(7x-2y)/ xy = 5

7/y â€“ 2/x = 5â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

SubstitutingÂ 1/x =mÂ in the given equation we get,

â€“ 2m + 7n = 5Â Â Â Â  => -2 + 7n â€“ 5 = 0Â  â€¦â€¦..(iii)

7m + 8n = 15Â Â Â Â  => 7m + 8n â€“ 15 = 0 â€¦â€¦(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

m = 1

n/(-65) = 1/(-65)

n = 1

m = 1 and n = 1

m =Â 1/xÂ = 1Â Â Â Â Â Â Â  n =Â 1/xÂ = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

Solution:

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

Â Â Â

2x + 4y = 5xy

=>Â 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

3m + 6n â€“ 6 = 0

4m + 2n â€“ 5 = 0

By cross-multiplication method, we get

m/(-30 â€“(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

m = 1

n/-9 = 1/-18

n = 1/2

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1Â andÂ y = 2

Hence, x = 1 and y = 2

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

15/(x+y) â€“ 5/(x-y) = -2

Solution:

SubstitutingÂ 1/x+y = mÂ andÂ 1/x-y = nÂ in the given equations, we get,

10m + 2n = 4Â Â Â Â Â  =>Â  10m + 2n â€“ 4 = 0Â Â Â Â Â  â€¦â€¦â€¦â€¦â€¦â€¦..â€¦..(i)

15m â€“ 5n = -2Â Â Â Â  =>Â Â  15m â€“ 5n + 2 = 0Â Â Â  â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

Using cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iii)

n = 1/(x-y) = 1

x-y = 1â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(iv)

Adding equation (iii) and (iv), we get

2x = 6Â Â  => x = 3 â€¦â€¦.(v)

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

y = 2

Hence, x = 3 and y = 2

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

1/2(3x+y) â€“ 1/2(3x-y) = -1/8

Solution:

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦â€¦ (1)

m/2 â€“ n/2 = -1/8

m â€“ n = -1/4Â  â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..â€¦(2)

Adding (1) and (2), we get

2m = 3/4 â€“ 1/4

2m = 1/2

Putting in (2), we get

1/4 â€“ n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4Â  â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(3)

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

3x â€“ y = 2 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(4)

Adding equations (3) and (4), we get

6x = 6

x = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(5)

Putting in (3), 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.

(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 us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu during,

Downstream = x + y km/h

Upstream = x â€“ y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

And, 2(x-y) = 4

Or x â€“ y = 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

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

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given,

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

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

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

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

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

3m + 6n =1/3 => 9m + 18n = 1â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore,

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

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

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

100/x + 200/y = 25/6 â€¦â€¦â€¦â€¦â€¦.(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(3)

100m + 200n = 25/6

600m + 1200n = 25 â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60

And y = 1/n

y = 80

Therefore,

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

The exercise after each topic helps the students evaluate their understanding of the topic. This exercise contains 2 questions divided into several parts. The solutions to these questions are provided in NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables. The steps to the solutions are accurate and are provided by the subject experts.

### Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.6

• The questions are solved by the subject experts.
• The answers to the questions are accurate.
• The solutions to all the questions in the textbook are provided here.
• NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.6 can help the students to score well in the first term examinations.