Pair of Linear Equations in Two Variables Exercise 3.6

NCERT solutions for Maths Pair of Linear Equations in Two Variables 

NCERT Solutions for Class 10 Maths Pair of Linear Equations in Two Variables

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

(i)

(ii)

(iii) + 3y = 14

− 4y = 23

(iv)

NCERT Solutions for Class 10 Maths Exercise 3.6

(v) 7x − 2y = 5xy

8x + 7y = 15xy

(vi) 6x + 3y − 6xy = 0

2x + 4y − 5xy = 0

(vii)

(viii)

Ans. (i) … (1)

… (2)

Let = p and = q

Putting this in equation (1) and (2), we get

⇒ 3p + 2q = 12 and 6 (2p + 3q) = 13 (6)

⇒ 3p + 2q = 12 and 2p + 3q =13

⇒ 3p + 2q – 12 = 0 … (3) and 2p + 3q – 13 = 0 … (4)

⇒

⇒

⇒

⇒ p = 2 and q = 3

But = p and = q

Putting value of p and q in this we get

x = and y =

(ii)… (1)

… (2)

Let = p and = q

Putting this in (1) and (2), we get

2p + 3q = 2 … (3)

4p − 9q = −1 … (4)

Multiplying (3) by 2 and subtracting it from (4), we get

4p − 9q + 1 – 2 (2p + 3q − 2) = 0

⇒ 4p − 9q + 1 − 4p − 6q + 4 = 0

⇒ −15q + 5 = 0

⇒ q =

Putting value of q in (3), we get

2p + 1 = 2 ⇒ 2p = 1⇒ p = ½

Putting values of p and q in (= p and = q), we get

and

⇒

⇒ x = 4 and y = 9

NCERT Solutions for Class 10 Maths Exercise 3.6

(iii)+ 3y = 14 … (1)

− 4y = 23 … (2) and Let = p … (3)

Putting (3) in (1) and (2), we get

4p + 3y = 14 … (4)

3p − 4y = 23 … (5)

Multiplying (4) by 3 and (5) by 4, we get

3 (4p + 3y – 14 = 0) and, 4 (3p − 4y – 23 = 0)

⇒ 12p + 9y – 42 = 0 … (6) 12p − 16y – 92 = 0 … (7)

Subtracting (7) from (6), we get

9y − (−16y) – 42 − (−92) = 0

⇒ 25y + 50 = 0

⇒ y = 50 – 25 = −2

Putting value of y in (4), we get

4p + 3 (−2) = 14

⇒ 4p – 6 = 14

⇒ 4p = 20⇒ p = 5

Putting value of p in (3), we get

= 5 ⇒ x =

Therefore, x = and y = −2

(iv) … (1)

… (2)

Let

Putting this in (1) and (2), we get

5p + q = 2

⇒ 5p + q – 2 = 0 … (3)

And, 6p − 3q = 1

⇒ 6p − 3q – 1 = 0 … (4)

Multiplying (3) by 3 and adding it to (4), we get

3 (5p + q − 2) + 6p − 3q – 1 = 0

⇒ 15p + 3q – 6 + 6p − 3q – 1 = 0

⇒ 21p – 7 = 0

⇒ p =

Putting this in (3), we get

5 () + q – 2 = 0

⇒ 5 + 3q = 6

⇒ 3q = 6 – 5 = 1

⇒ q =

NCERT Solutions for Class 10 Maths Exercise 3.6

Putting values of p and q in (), we get

⇒ 3 = x − 1 and 3 = y – 2

⇒ x = 4 and y = 5

(v) 7x − 2y = 5xy … (1)

8x + 7y = 15xy … (2)

Dividing both the equations by xy, we get

Let = p and = q

Putting these in (3) and (4), we get

7q − 2p = 5 … (5)

8q + 7p = 15 … (6)

From equation (5),

2p = 7q – 5

⇒ p =

Putting value of p in (6), we get

8q + 7 () = 15

⇒ 16q + 49q – 35 = 30

⇒ 65q = 30 + 35 = 65

⇒ q = 1

Putting value of q in (5), we get

7 (1) − 2p = 5

⇒ 2p = 2⇒ p = 1

NCERT Solutions for Class 10 Maths Exercise 3.6

Putting value of p and q in (= p and = q), we get x = 1 and y = 1

(vi) 6x + 3y − 6xy = 0 … (1)

2x + 4y − 5xy = 0 … (2)

Dividing both the equations by xy, we get

Let = p and = q

Putting these in (3) and (4), we get

6q + 3p – 6 = 0 … (5)

2q + 4p – 5 = 0 … (6)

From (5),

3p = 6 − 6q

⇒ p = 2 − 2q

Putting this in (6), we get

2q + 4 (2 − 2q) – 5 = 0

⇒ 2q + 8 − 8q – 5 = 0

⇒ −6q = −3⇒ q = ½

Putting value of q in (p = 2 – 2q), we get

p = 2 – 2 (½) = 2 – 1 = 1

NCERT Solutions for Class 10 Maths Exercise 3.6

Putting values of p and q in (= p and = q), we get x = 1 and y = 2

(vii) … (1)

…(2)

Let

Putting this in (1) and (2), we get

10p + 2q = 4 … (3)

15p − 5q = −2 … (4)

From equation (3),

2q = 4 − 10p

⇒ q = 2 − 5p … (5)

Putting this in (4), we get

15p – 5 (2 − 5p) = −2

⇒ 15p – 10 + 25p = −2

⇒ 40p = 8⇒ p =

Putting value of p in (5), we get

q = 2 – 5 () = 2 – 1 = 1

Putting values of p and q in (), we get

⇒ x + y = 5 … (6) and x – y = 1 … (7)

Adding (6) and (7), we get

2x = 6 ⇒ x = 3

Putting x = 3 in (7), we get

3 – y = 1

⇒ y = 3 – 1 = 2

Therefore, x = 3 and y = 2

NCERT Solutions for Class 10 Maths Exercise 3.6

(viii) … (1)

… (2)

Let

Putting this in (1) and (2), we get

p + q = and

⇒ 4p + 4q = 3 … (3) and 4p − 4q = −1 … (4)

Adding (3) and (4), we get

8p = 2 ⇒ p = ¼

Putting value of p in (3), we get

4 (¼) + 4q = 3

⇒ 1 + 4q = 3

⇒ 4q = 3 – 1 = 2

⇒ q = ½

Putting value of p and q in , we get

⇒ 3x + y = 4 … (5) and 3x – y = 2 … (6)

Adding (5) and (6), we get

6x = 6 ⇒ x = 1

Putting x = 1 in (5) , we get

3 (1) + y = 4

⇒ y = 4 – 3 = 1

Therefore, x = 1 and y = 1

NCERT Solutions for Class 10 Maths Exercise 3.6

2. Formulate the following problems as a part 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.

Ans. (i) Let speed of rowing in still water = x km/h

Let speed of current = y km/h

So, speed of rowing downstream = (x + y) km/h

And, speed of rowing upstream = (x − y) km/h

According to given conditions,

⇒ 2x + 2y = 20 and 2x − 2y = 4

⇒ x + y = 10 … (1) and x – y = 2 … (2)

Adding (1) and (2), we get

2x = 12⇒ x = 6

Putting x = 6 in (1), we get

6 + y = 10

⇒ y = 10 – 6 = 4

Therefore, speed of rowing in still water = 6 km/h

Speed of current = 4 km/h

NCERT Solutions for Class 10 Maths Exercise 3.6

(ii) Let time taken by 1 woman alone to finish the work = x days

Let time taken by 1 man alone to finish the work = y days

So, 1 woman’s 1-day work = ()th part of the work

And, 1 man’s 1-day work = ()th part of the work

So, 2 women’s 1-day work = ()th part of the work

And, 5 men’s 1-day work = ()th part of the work

Therefore, 2 women and 5 men’s 1-day work = (+)th part of the work… (1)

It is given that 2 women and 5 men complete work in = 4 days

It means that in 1 day, they will be completing th part of the work … (2)

Clearly, we can see that (1) = (2)

⇒ … (3)

Similarly, … (4)

Let

Putting this in (3) and (4), we get

2p + 5q = and 3p + 6q =

⇒ 8p + 20q = 1 … (5) and 9p + 18q = 1 … (6)

Multiplying (5) by 9 and (6) by 8, we get

72p + 180q = 9 … (7)

72p + 144q = 8 … (8)

Subtracting (8) from (7), we get

36q = 1⇒ q =

Putting this in (6), we get

9p + 18 () = 1

⇒ 9p = ½⇒ p =

Putting values of p and q in , we get x = 18 and y = 36

Therefore, 1 woman completes work in = 18 days

And, 1 man completes work in = 36 days

NCERT Solutions for Class 10 Maths Exercise 3.6

(iii) Let speed of train = x km/h and let speed of bus = y km/h

According to given conditions,

Let

Putting this in the above equations, we get

60p + 240q = 4 … (1)

And 100p + 200q = … (2)

Multiplying (1) by 5 and (2) by 3, we get

300p + 1200q = 20 … (3)

300p + 600q = … (4)

Subtracting (4) from (3), we get

600q = 20 − = 7.5

⇒ q =

Putting value of q in (2), we get

100p + 200 () =

⇒ 100p + 2.5 =

⇒ 100p = – 2.5

⇒ p =

But

Therefore, x =km/h and y = km/h

Therefore, speed of train = 60 km/h

And, speed of bus = 80 km/h