Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

The document Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Ques 1: Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(ii) 
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(iii) 
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(iv) 
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev 
(v) 
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev 
(vi) 
6x + 3y = 6xy 
2x + 4y = 5xy
(vii) Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev 

(viii) Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Sol: (i) Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(1)
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
Multipliying (1) by 1/3 and (2) by 1/2, we get
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(3)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev   ...(4)
Subtracting (3) from (4),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Substituting v = 3 in (3), we get
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, v = 3 and u = 2
But Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ The required solution is  Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(ii) We have
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev   ...(1)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ (1) and (2) can be written as
2u + 3v = 2       ...(3)
4u − 9v = − 1     ...(4)
From (3) and (4),
a1 = 2, b1 = 3, c1 = - 2
a2 = 4, b2 = - 9, c2 = 1
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
But Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
The required solution is: Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(iii) We have:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev   ...(1)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
The equations (1) and (2) becomes,
4p + 3y = 14    ...(3)
3p − 4y = 23   ...(4)
Solving (3) and (4) by cross multiplication method:
a1 = 4, b1 = 3, c1 = - 14
a2 = 3, b2 = - 4, c2 = - 23
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Since, p = 1/x
∴ 1/x = 5 ⇒ x = 1/5
Thus, the required solution is: Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(iv) We have:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(1)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ Equations (1) and (2) can be expressed as:
5u + v = 2    ...(3)
6u - 3v = 1   ...(4)
Solving (3) and (4) by cross multiplication method:
a1 = 5, b1 = 1, c1 = - 2
a2 = 6, b2 = - 3, c2 = - 1
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
But Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ 3= x − 1 ⇒ x = 4
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ 3 = y − 2 ⇒ y = 5
Thus, the required solution is Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(v) We have:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(1)

Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
From equation (1),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(3)
From equation (2),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(4)
Let = 1/x = p and 1/y = q
∴ Equations (3) and (4) can be expressed as:
7q - 2p - 5 = 0
8q + 7p - 15 = 0
Using cross multiplication method to solve (5) and (6), we have:
a1 = 7, b1 = - 2, c1 = - 5
a2 = 8, b2 = 7, c2 = - 15
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Since, Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, the required solution is: Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(vi) We have:
6x + 3y = 6 xy   ...(1)
2x + 4y = 5 xy   ...(2)
From (1), we get
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(3)
From (2), we get
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(4)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ (3) and (4) can be expressed as
6q + 3p = 6 ...(5)
2q + 4p = 5 ...(6)
Multiply (6) by 3 and subtracting from (5),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ p = -9/-9 = 1
Substituting p = 1 in (5), we get
6q + 3 (1) = 6
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Since, Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, the required solution is Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(vii) We have:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(1)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(2)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ Equations (1) and (2) can be expressed as
10p + 2q = 4     ...(3)
15p − 5q = − 2    ...(4)
To solve (3) and (4),
a1 = 10, b1 = 2, c1 = - 4
a2 = 15, b2 = - 5, c2 = 2
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
and Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
But Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒   x + y = 5   ...(5)
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev ...(6)
Adding (5) and (6), we have
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
From (5), 3 + y = 5 ⇒ y = 5 − 3 = 2
Thus, the required solution is: Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
(viii) We have:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(1)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(2)
Let Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ (1) and (2) can be expressed as:
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev     ...(3)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(4)
Multiplying the equation (3) by 1/2 and adding it to (4),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
From (3), Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
But Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ 3x + y = 4    ...(5)
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

⇒ 3x − y = 2 ...(6)
Adding (5) and (6),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Subtracting (6) from (5),
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, the required solution is: Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Ques 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.
Sol: (i) Let the speed of Ritu in still water = x km/hr
And the speed of the water current = y km/hr
∴ Downstream speed = (x + y) km/hr
Upstream speed = (x - y) km/hr
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev                   
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x + y = 20/2
⇒ x + y = 10    ...(1)
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x - y = 4/2
⇒ x - y = 2       ... (2)
Adding (1) and (2), we get
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x = 12/2 = 6
from, (1) 6 + y = 10
⇒ y = 10 − 6 = 4
Thus, speed of Ritu = 6 km/hr
speed of water current = 4 km/hr
(ii) Let the time taken to finish the task
by one woman alone = x days
by one man alone = y days
∴1 woman’s 1 day work = 1/x
1 man’s 1 day work = 1/y
Since, [2 women + 5 man], finish the task in 4 days
∴  Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒   Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev   ...(1)
Again
[3 women + 6 men], finish the task in 3 days
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev      ...(2)
Let 1/x = p and 1/y = q
∴ Equation (1) and (2) are expressed as
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Using cross multiplication method, we get
a1 = 8,  b1 = 20,  c1 = - 1
a2 = 9,  b2 = 18,  c2 = - 1
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Since, Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
And Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, required number of day
for a woman = 18 days
for a man = 36 days
(iii) Let the speed of the train = x km/hr
And the speed of the bus = y km/hr
Since, time = distance/speed
Case I:
Total Journey = 300 km
∵ Journey travelled by train = 60 km
∴ Journey travelled by bus = (300 - 60) km
= 240 km
∵ Total time taken = 4 hours
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(1) [Dividing by 15]
Case II:
Distance travelled by train = 100 km
∴ Distance travelled by bus = (300 - 100) km
= 200 km
Total time = 4 hrs 10 mts
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev ...(2) [Dividing by 25]
Multiply (1) by 4 and subtracting from (2)
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = 80
From (1), Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.6 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
∴ x = 60
Thus, speed of the train = 60 km/hr
speed of the bus = 80 km/hr

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