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NCERT Solutions for Class 10 Maths Exercise 3.6 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**NCERT solutions for Maths Pair of Linear Equations in Two Variables Download as PDF**

## 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) + 3 y = 14**

** − 4 y = 23**

**(iv) **

NCERT Solutions for Class 10 Maths Exercise 3.6

**(v) 7 x − 2y = 5xy **

**8 x + 7y = 15xy**

**(vi) 6 x + 3y − 6xy = 0**

**2 x + 4y − 5xy = 0**

**(vii) **

**(viii) **

**Ans. (i)** … (1)

… (2)

Let = *p* and = *q*

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

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

⇒ 3*p *+ 2*q *= 12 and 2p + 3q =13

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

2*p *+ 3*q *= 2 … (3)

4*p *− 9*q *= −1 … (4)

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

4*p *− 9*q *+ 1 – 2 (2*p *+ 3*q *− 2) = 0

⇒ 4*p *− 9*q *+ 1 − 4*p *− 6*q *+ 4 = 0

⇒ −15*q *+ 5 = 0

⇒ *q *=

Putting value of q in (3), we get

2*p *+ 1 = 2 ⇒ 2*p *= 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)** + 3*y *= 14 … (1)

− 4*y *= 23 … (2) and Let = *p … *(3)

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

4*p *+ 3*y *= 14 … (4)

3*p *− 4*y *= 23 … (5)

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

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

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

Subtracting (7) from (6), we get

9*y *− (−16*y*) – 42 − (−92) = 0

⇒ 25*y *+ 50 = 0

⇒ *y *= 50 – 25 = −2

Putting value of y in (4), we get

4*p *+ 3 (−2) = 14

⇒ 4*p *– 6 = 14

⇒ 4*p *= 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

5*p *+ *q *= 2

⇒ 5*p *+ *q *– 2 = 0 … (3)

And, 6*p *− 3*q *= 1

⇒ 6*p *− 3*q *– 1 = 0 … (4)

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

3 (5*p *+ *q *− 2) + 6*p *− 3*q *– 1 = 0

⇒ 15*p *+ 3*q *– 6 + 6*p *− 3*q *– 1 = 0

⇒ 21*p *– 7 = 0

⇒ *p *=

Putting this in (3), we get

5 () + *q *– 2 = 0

⇒ 5 + 3*q *= 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)** 7*x *− 2*y *= 5*xy … *(1)

8*x *+ 7*y *= 15*xy …* (2)

Dividing both the equations by xy, we get

Let = *p* and = *q*

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

7*q *− 2*p *= 5 … (5)

8*q *+ 7*p *= 15 … (6)

From equation (5),

2*p *= 7*q *– 5

⇒ *p *=

Putting value of p in (6), we get

8*q *+ 7 () = 15

⇒ 16*q *+ 49*q *– 35 = 30

⇒ 65*q *= 30 + 35 = 65

⇒ *q *= 1

Putting value of q in (5), we get

7 (1) − 2*p *= 5

⇒ 2*p *= 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)** 6*x *+ 3*y *− 6*xy *= 0 … (1)

2*x *+ 4*y *− 5*xy *= 0 … (2)

Dividing both the equations by xy, we get

Let = *p* and = *q*

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

6*q *+ 3*p *– 6 = 0 … (5)

2*q *+ 4*p *– 5 = 0 … (6)

From (5),

3*p *= 6 − 6*q*

⇒ *p *= 2 − 2*q*

Putting this in (6), we get

2*q *+ 4 (2 − 2*q*) – 5 = 0

⇒ 2*q *+ 8 − 8*q *– 5 = 0

⇒ −6*q *= −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

10*p *+ 2*q *= 4 … (3)

15*p *− 5*q *= −2 … (4)

From equation (3),

2*q *= 4 − 10*p*

⇒ *q *= 2 − 5*p … *(5)

Putting this in (4), we get

15*p *– 5 (2 − 5*p*) = −2

⇒ 15*p *– 10 + 25*p *= −2

⇒ 40*p *= 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

2*x *= 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

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

Adding (3) and (4), we get

8*p *= 2 ⇒ *p *= ¼

Putting value of p in (3), we get

4 (¼) + 4*q *= 3

⇒ 1 + 4*q *= 3

⇒ 4*q *= 3 – 1 = 2

⇒ *q *= ½

Putting value of p and q in , we get

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

Adding (5) and (6), we get

6*x *= 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,

⇒ 2*x *+ 2*y *= 20 and 2*x *− 2*y *= 4

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

Adding (1) and (2), we get

2*x *= 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

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

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

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

72*p *+ 180*q *= 9 … (7)

72*p *+ 144*q *= 8 … (8)

Subtracting (8) from (7), we get

36*q *= 1⇒ *q *=

Putting this in (6), we get

9*p *+ 18 () = 1

⇒ 9*p *= 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

60*p *+ 240*q *= 4 … (1)

And 100*p *+ 200*q *= … (2)

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

300*p *+ 1200*q *= 20 … (3)

300*p *+ 600*q *= … (4)

Subtracting (4) from (3), we get

600*q *= 20 − = 7.5

⇒ *q *=

Putting value of q in (2), we get

100*p *+ 200 () =

⇒ 100*p *+ 2.5 =

⇒ 100*p *= – 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

## NCERT Solutions for Class 10 Maths Exercise 3.6

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