**NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations In Two Variables**

NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations are created here for boosting your preparation of the exam.NCERT Solutions for Class 10 Maths Exercise 3.6 of Chapter 3 Pair of Linear Equations will clear your basic concepts on linear equations in two variables. Exercise 3.6 has different structures of linear equations which are solved here in the easiest ways. These NCERT solutions are not only helpful to the class 10 maths students but also for the candidates who are going to appear in the competitive exams.

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**NCERT S**olutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

**Exercise 3.1 -Linear Equations in Two Variables**

**Exercise 3.2 -Linear Equations in Two Variables**

**Exercise 3.3-Linear Equation in Two Variables**

**Exercise 3.4-Pair of Linear Equations**

**Exercise 3.5-Pair of Linear Equations**

**Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations**

**Exercise 3.7 – Linear Equations in Two Variables**

**Latest Sample paper Class 10 maths for Term 1 2021 CBSE board**

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

**Q1. 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**

Ans.The given pair of linear equations is

**1/2x + 1/3y = 2…..(i)**

**1/3x + 1/2y = 13/6….(ii)**

Let 1/x = a and 1/y = b,then we get equation (iii) and equation (iv)

a/2 + b/3 = 2

3a + 2b =12…….(iii)

a/3 + b/2 =13/6

2a + 3b =13/6

12a +18b = 78….(iv)

Multiplying equation (iii) by 4,we get equation (v)

12a + 8b = 48……(v)

Substracting equation (v) from equation (iv)

10b = 30

b = 3

Putting the value of b in equation (iii)

3a + 2×3 = 12

3a +6 =12

3a = 12-6 =6

a = 2

Since 1/x = a and 1/y = b

Substituting the value a and b ,we get x = 1/2 and y = 1/3

**(ii) The given pair of linear equation is**

**2/√x + 3/√y = 2……(i)**

**4/√x – 9/√y = -1…..(ii)**

Let 1/√x = a and b = 1/√y, we get the equations (iii) and equation (iv)

2a + 3b = 2…….(iii)

4a – 9b =-1……(iv)

Multiplying equation (iii) by 2,we get equation (v)

4a + 6b = 4…..(v)

Substracting equation (v) from equation (iv)

-15b = -5

b = 1/3

Putting the value of b in equation (iii)

2a + 3(1/3) =2

2a +1 =2

2a =1

a =1/2

Since 1/√x = a and 1/√y = b

Substituting the value of a and b

1/√x = 1/2 and 1/√y = 1/3

Squaring both equations

1/x = 1/4 and 1/y = 1/9

x =4 and y = 9

**(iii) The given equations are **

**4/x + 3y = 14….(i)**** **

**3/x -4y = 23……(ii)**

Let 1/x =z

4z + 3y =14……(iii)

3z – 4y =23……(iv)

Multiplying equation (iii) by 3 and equation (iv) by 4,we get equation (v) and equation (vi)

12z + 9y = 42…..(v)

12z – 16y = 92……(vi)

Substracting equation (vi) from equation (v)

25 y = -50

y = -2

Putting the value of y in equation (v)

12z +9×-2 =42

12z -18 =42

12z = 42 +18 =60

z = 5

Since 1/x = z

Substituting the value of z

x = 1/5

Therefore the value of x is 1/5 and the value of y is -2

**(iv) The given equations are **

** 5/(x-1) + 1/(y-2) = 2…..(i)**

**6/(x-1) – 3/(y-2) = 1….(ii)**

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

5a + b = 2……(iii)

6a – 3b = 1……(iv)

Multiplying equation (iii) by 3,we get equation (v)

15a + 3b = 6….(v)

Adding equation (iv) and (v)

21a = 7

a = 7/21 = 1/3

Putting the value of a in equation (iii)

5×1/3 +b =2

5/3 + b =2

b = 2 -(5/3)

b = 1/3

Putting the value of a and b in 1/(x -1) = a and 1/(y-2) = b

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

x -1 =3 and y-2 =3

x = 4 and y =5

**(v) The given pair of linear equations is**

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

**(8x + 7y)/xy = 15**

Simplifying the given equations

7x/xy – 2y/xy = 5

7/y – 2/x = 5……(i)

8x/xy +7y/xy = 15

8/y +7/x = 15….(ii)

Let 1/y =a and 1/x = b

7a -2b = 5……(iii)

8a +7b = 15…..(iv)

Multiplying equation (iii) by 8 and equation (iv) by 7,we get equation (v) and equation (vi)

56a – 16b = 40….(v)

56a + 49b = 105…(vi)

Substracting equation (vi) from equation (v)

-65b = -65

b = 1

Putting the value of b in equation (iii)

7a -2×1 = 5

7a -2 = 5

7a = 7

a = 1

Since 1/y =a and 1/x = b

∴ 1/y = 1 and 1/x = 1

y = 1 and x = 1

**(vi) The given equations are **

**6x + 3y = 6xy**

**2x + 4y = 5xy**

Simplifying the given equations

6x/xy +3y/xy = 6xy/xy

6/y + 3/x = 6….(i)

2x/xy + 4y/xy = 5xy/xy

2/y + 4/x = 5…(ii)

Let 1/y =a and 1/x = b

6a +3b = 6……(iii)

2a + 4b =5….(iv)

Multiplying equation (iii) by 2 and equation (iv) by 6,we get equation (v) and equation (vi)

12a +6b = 12….(v)

12a + 24b = 30….(vi)

Substracting equation (vi) from equation (v)

-18b = -18

b = 1

Putting the value of b in equation (iii)

6a + 3×1=6

6a + 3 = 6

6a = 3

a = 3/6 = 1/2

Since 1/y =a and 1/x = b

Substituting the value of a and b

1/y =1/2 and 1/x = 1

y =2 and x =1

**(vii) The given pair of linear equations is**

**10/(x+y) + 2/(x-y) = 4…..(i)**

**15/(x+y) – 5/(x-y) = -2….(ii)**

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

10a + 2b = 4…….(iii)

15a – 5b = = -2….(iv)

Multiplying equation (iii) by 5 and equation (iv) by 2,we get equation (v) and equation (vi)

50a + 10b = 20…….(v)

30a – 10b = = -4….(vi)

Adding both of the equation (v) and equation (vi)

80a = 16

a = 16/80 = 1/5

Putting the value of a in equation (v)

50 ×1/5 +10b =20

10 + 10b =20

10b = 20-10 = 10

b = 1

Since 1/(x+y) = a and 1/(x-y) = b

Substituting the value of a and b ,we get equation (vii) and equation (viii)

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

x + y = 5…..(vii) and x -y =1(viii)

Adding both equations (vii) and (viii)

2x = 6

x = 3

Putting the value of x in equation (vii)

3 + y = 5

y = 5-3 = 2

**Hence the value of x is 3 and value of y is 2**

**(viii) 1/(3x+y) + 1/(3x-y) = 3/4….(i)**

**1/2(3x+y) – 1/2(3x-y) = -1/8…..(ii)**

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

a + b = 3/4

4a + 4b = 3…..(iii)

a/2 – b/2 = -1/8

4a – 4b =-1…..(iv)

Adding both equations (iii) and (iv)

8a = 2

a = 2/8 = 1/4

Putting the value of a in equation (iv)

4×1/4 + 4b =3

1 + 4b = 3

4b = 3-1 = 2

b = 2/4 = 1/2

Since 1/(3x+y) =a and 1/(3x-y) = b

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

3x + y = 4 ….(v) and 3x -y = 2….(vi)

Adding the equations (v) and equation (vi)

6x = 6

x = 1

Putting the value of x in equation (v)

3 ×1+ y =4

3 + y = 4

y = 4 -3 = 1

Hence the value of x is and value of y is 1

**Q2.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.**

Ans. Let the speed of the boat in still water is x km/h and the speed of the current is y km/h

The speed of boat in down stream is = x + y

Since we are given distance covered by the boat in downstream is 20 km in time 2 hour

Therefore speed of boat in downstream = Distance/Time = 20/2 = 10 km/h

According to first condition

x + y = 10…….(i)

The speed of boat in down stream is = x – y

Since we are given distance covered by the boat in upstream is 4 km in time 2 hour

Therefore speed of boat in upstream = Distance/Time = 4/2 = 2 km/h

According to second condition

x – y = 2……(ii)

Adding both equations (i) and (ii)

2x = 8

x = 8/2 = 4

Putting the value of x in equation (ii)

4 -y = 2

y = 4 -2 = 2

Hence the speed of the boat in still water is 4 km/h and the speed of the current is 2 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.**

Ans.Let 1 woman alone can finish the work in x days and 1 man alone can finish the work in y days

Since in x days 1 woman do 1 work

In 1 day a woman will do = 1/x work

In 1 day 2 women will do = 2/x work

Since in y days, 1 man do 1 work

In 1-day a man will do = 1/y work

In 1 day 5 men will do = 5/y work

2 woman and 5 men do a work in 4 days

In 1 day they will work =1/4 work

Therefore according to first condition

2/x + 5/y = 1/4….(i)

Similarly according to second condition,we can creat second equation

3/x + 6/y = 1/3….(ii)

Let 1/x =a and 1/y = b

2a + 5b = 1/4

8a + 20b =1…….(iii)

3a +6b =1/3

9a + 18b =1……(iv)

Multiplying equation (iii) by 9 and equation (iv) by 8

72a + 180b = 9…(v)

72a +144b =8….(vi)

36b = 1

b = 1/36

Putting the value of b in equation (iii)

8a + 20×1/36 =1

8a + 5/9 = 1

8a = 1-5/9 = 4/9

a = 1/18

Since 1/x =1/18 and 1/y = 1/36

x =18 and y = 36

Hence a woman alone can do the work in 18 days and a man alone can do the given work in 36 days.

**(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.**

Ans. Let the speed of the train is x km/h and speed of the bus is y km/h

According to first condition Roohi travels 60 km by train and the remaining (300 -60 = 240 km) by bus and total time taken is 4 hours.

Time taken to cover the distance 60 km + Time taken to cover the distance 240 km/h = 4 hours

Since time = Distance/Speed

60/x + 240/y = 4……(i)

According to second condition Roohi travels 100 km by train and the remaining (300 -100 = 200 km) by bus and total time taken is 4 hours +10 minutes

Time taken to cover the distance 100 km + Time taken to cover the distance 200 km/h = (4+10/60)=4+1/6 =(25/6)h

Since time = Distance/Speed

100/x + 200/y = 25/6……(ii)

Let a = 1/x and b = 1/y

60a + 240b = 4…….(iii)

100a + 200b = 25/6

600a + 1200b = 25….(iv)

Multiplying the equation (iii) by 10,we get equation (v)

600a + 2400b = 40…(v)

Substracting equation (v) from equation (iv)

-1200b = -15

b = 15/1200 = 1/80

Putting the value of b in equation

60a + 240×1/80 =4

60a +3 = 4

60a = 4-3 =1

a = 1/60

Since a = 1/x and b = 1/y

1/60 = 1/x and 1/80 = 1/y

x = 60 and y =80

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

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