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

## 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)

(ii)

(iii)

(iv)

(v)

(vi)
6x + 3y = 6xy
2x + 4y = 5xy
(vii)

(viii)

Sol: (i) Let
âˆ´     ...(1)
And   ...(2)
Multipliying (1) by 1/3 and (2) by 1/2, we get
...(3)
...(4)
Subtracting (3) from (4),

â‡’
â‡’
â‡’
Substituting v = 3 in (3), we get

â‡’
â‡’
Thus, v = 3 and u = 2
But
And
âˆ´ The required solution is
(ii) We have
...(1)
...(2)
Let
âˆ´ (1) and (2) can be written as
2u + 3v = 2       ...(3)
4u âˆ’ 9v = âˆ’ 1     ...(4)
From (3) and (4),
a1 = 2, b1 = 3, c1 = - 2
a2 = 4, b2 = - 9, c2 = 1
âˆ´
â‡’
â‡’
â‡’

But
âˆ´

The required solution is:
(iii) We have:
...(1)
...(2)
Let
The equations (1) and (2) becomes,
4p + 3y = 14    ...(3)
3p âˆ’ 4y = 23   ...(4)
Solving (3) and (4) by cross multiplication method:
a1 = 4, b1 = 3, c1 = - 14
a2 = 3, b2 = - 4, c2 = - 23
âˆ´

â‡’
â‡’
âˆ´
And
Since, p = 1/x
âˆ´ 1/x = 5 â‡’ x = 1/5
Thus, the required solution is:
(iv) We have:
...(1)
...(2)
Let
âˆ´ 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
âˆ´
â‡’
â‡’
â‡’
âˆ´

But
â‡’ 3= x âˆ’ 1 â‡’ x = 4
And
â‡’ 3 = y âˆ’ 2 â‡’ y = 5
Thus, the required solution is
(v) We have:
...(1)

...(2)
From equation (1),

â‡’     ...(3)
From equation (2),

â‡’     ...(4)
Let = 1/x = p and 1/y = q
âˆ´ 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
âˆ´

â‡’
â‡’
âˆ´

Since,
And
Thus, the required solution is:
(vi) We have:
6x + 3y = 6 xy   ...(1)
2x + 4y = 5 xy   ...(2)
From (1), we get
...(3)
From (2), we get
...(4)
Let
âˆ´ (3) and (4) can be expressed as
6q + 3p = 6 ...(5)
2q + 4p = 5 ...(6)
Multiply (6) by 3 and subtracting from (5),

â‡’ p = -9/-9 = 1
Substituting p = 1 in (5), we get
6q + 3 (1) = 6
â‡’
Since,
And
Thus, the required solution is
(vii) We have:
...(1)
...(2)
Let
âˆ´ Equations (1) and (2) can be expressed as
10p + 2q = 4     ...(3)
15p âˆ’ 5q = âˆ’ 2    ...(4)
To solve (3) and (4),
a1 = 10, b1 = 2, c1 = - 4
a2 = 15, b2 = - 5, c2 = 2
âˆ´

â‡’
â‡’
âˆ´
and
But
â‡’   x + y = 5   ...(5)
And  ...(6)
Adding (5) and (6), we have

From (5), 3 + y = 5 â‡’ y = 5 âˆ’ 3 = 2
Thus, the required solution is:
(viii) We have:
...(1)
...(2)
Let
âˆ´ (1) and (2) can be expressed as:
...(3)
...(4)
Multiplying the equation (3) by 1/2 and adding it to (4),

From (3),
But
â‡’ 3x + y = 4    ...(5)
And

â‡’ 3x âˆ’ y = 2 ...(6)

Subtracting (6) from (5),

Thus, the required solution is:

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
âˆ´ Downstream speed = (x + y) km/hr
Upstream speed = (x - y) km/hr
âˆµ
âˆ´
â‡’ x + y = 20/2
â‡’ x + y = 10    ...(1)
And
â‡’ x - y = 4/2
â‡’ x - y = 2       ... (2)
Adding (1) and (2), we get

â‡’ x = 12/2 = 6
from, (1) 6 + y = 10
â‡’ y = 10 âˆ’ 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
âˆ´1 womanâ€™s 1 day work = 1/x
1 manâ€™s 1 day work = 1/y
Since, [2 women + 5 man], finish the task in 4 days
âˆ´
â‡’      ...(1)
Again
[3 women + 6 men], finish the task in 3 days
âˆ´
â‡’       ...(2)
Let 1/x = p and 1/y = q
âˆ´ Equation (1) and (2) are expressed as

Using cross multiplication method, we get
a1 = 8,  b1 = 20,  c1 = - 1
a2 = 9,  b2 = 18,  c2 = - 1
âˆ´
â‡’
â‡’
âˆ´

Since,
And
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
âˆµ Journey travelled by train = 60 km
âˆ´ Journey travelled by bus = (300 - 60) km
= 240 km
âˆµ Total time taken = 4 hours
âˆ´
â‡’   ...(1) [Dividing by 15]
Case II:
Distance travelled by train = 100 km
âˆ´ Distance travelled by bus = (300 - 100) km
= 200 km
Total time = 4 hrs 10 mts

âˆ´
â‡’  ...(2) [Dividing by 25]
Multiply (1) by 4 and subtracting from (2)

â‡’
â‡’ y = 80
From (1),
â‡’
â‡’
âˆ´ x = 60
Thus, speed of the train = 60 km/hr
speed of the bus = 80 km/hr

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