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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) **

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),

a

a

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:

a

a

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:

a

a

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:

a_{1} = 7, b_{1} = - 2, c_{1} = - 5

a_{2} = 8, b_{2} = 7, c_{2} = - 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),

a_{1} = 10, b_{1} = 2, c_{1} = - 4

a_{2} = 15, b_{2} = - 5, c_{2} = 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)

Adding (5) and (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

a_{1} = 8, b_{1} = 20, c_{1} = - 1

a_{2} = 9, b_{2} = 18, c_{2} = - 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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