Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1.

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

(i) \(\frac{1}{2 x}\) + \(\frac{1}{3 y}\) = 2

\(\frac{1}{3 x}\) + \(\frac{1}{2 y}\) = \(\frac{13}{6}\)

(ii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2

\(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1

(iii) \(\frac{4}{x}\) + 3y = 14

\(\frac{3}{x}\) – 4y = 23

(iv) \(\frac{5}{x-1}\) + \(\frac{1}{y-1}\) = 2

\(\frac{6}{x-1}\) – \(\frac{3}{y-2}\) = 1

(v) \(\frac{7 x-2 y}{x y}\) = 5

\(\frac{8x+7y}{x y}\) = 15

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

(vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4

\(\frac{15}{x+y}\) + \(\frac{5}{x-y}\) = -2

(viii) \(\frac{1}{3 x+y}\) + \(\frac{1}{3 x-y}\) = \(\frac{3}{4}\)

\(\frac{1}{2(3 x+y)}\) – \(\frac{1}{2(3 x-y)}\) = \(\frac{-1}{8}\)

Solution:

(i) The given pair of linear equations is :

3u + 2v = 12 …(1)

and \(\frac{u}{3}+\frac{v}{2}\) = \(\frac{13}{6}\)

or \(\frac{2 u+3 v}{6}\) = \(\frac{13}{6}\)

or 2u + 3v = 13 ….(2)

Multiplying equation (1) by 2 and (2) by 3,

6u + 4v = 24 …(3)

and 6u + 9v = 39 …(4)

Subtracting (3) from equation (4),

Substituting this value of v in (1),

3u + 2(3) = 12

or 3u + 6 = 12

or 3u = 12 – 6 = 6

or u = \(\frac{6}{3}\) = 2

2u + 3v = 2 ….(1)

and 4u – 9v = – 1 ….(2)

Multiplying equation (1) by 2

4u + 6v = 4 ….(3)

Subtracting (3) from equation (2),

Hence x = 4 and y = 9

(iii) The given pair of linear equations is

\(\frac{4}{x}\) + 3y = 14

and \(\frac{3}{x}\) – 4y = 23

Putting \(\frac{1}{x}\) = v

4v + 3y = 14 …(1)

and 3v – 4y = 23 …. (2)

Multiplying equation (1) by 3 and (2) by 4

12v + 9y = 42 …(3)

and 12v – 16y = 92

Subtracting (3) from (4)

Substituting this value of y in equation (1),

4v + 3 (-2) = 14

or 4v – 6 = 14

or 4v = 14 + 6 = 20

or v = \(\frac{20}{4}\) = 5

But \(\frac{1}{x}\) = v

or x = \(\frac{1}{5}\) and y = -2

Hence, x = \(\frac{1}{5}\) and y = – 2

(iv) The given pair of linear equations is

5u + v = 2 …(1)

6u – 3v = 1 …(2)

Multiplying equation (1) by 3

15u + 3v = 6

Adding equation (3) and (2)

15u + 3v = 6

6u – 3v = 1

21u = 7

⇒ u = \(\frac{7}{21}\) = \(\frac{1}{3}\)

Substituting this value of u in equation (1),

Hence, x = 4 and y = 5

(v) The given pair of linear equations is :

-2u + 7v = 5 …(1)

and 7u + 8v =15 ….(2)

Multiplying (1) by 7 and (2) by 2,

-14v + 49u = 35 ….(3)

and 14v + 16 u = 30 ….(4)

Adding equation (3) and (4)

-14v + 49u = 35

14v + 16u = 30

65u = 65

u = \(\frac{65}{65}\) = 1

Substituting this value of u in equation (1)

– 2 (1) + 7v = 5

or 7v = 5 + 2

or 7v = 7

or v = \(\frac{7}{7}\) = 1

Hence, x = 1 and y = 1

(vi) The given pair of linear equations is :

u + 2v = 2 ….(1)

and 4u + 2v = 5 ….(2)

Subtracting (1) from equation (2)

Substituting this value of u in (1)

1 + 2v = 2

or 2v = 2 – 1 = 1

or v = \(\frac{1}{2}\)

10u + 2v = 4

or 5u + v = 2 ….(1)

15u – 5v = – 2 ….(2)

Multiplying (1) by 5

25u + 5v = 10 ….(3)

Adding equation (3) and (2)

25u + 5v = 10

15u – 5v = -2

40u = 8

u = \(\frac{8}{40}\) = \(\frac{1}{5}\)

Substituting 4 in (1)

5 (\(\frac{1}{5}\)) + v = 2

or 1 + v = 2

or v = 1

Subtituting this value of x in (4)

3 + y = 5

or y = 5 – 3 = 2

Hence x = 3 and y = 2

(viii) The given pair of linear equations is :

u + v = \(\frac{3}{4}\)

or 4u + 4v = 3

or 4u + 4v = 3 ….(1)

and \(\frac{u}{2}\) – \(\frac{v}{2}\) = \(\frac{-1}{8}\)

or u – v = \(\frac{-1}{4}\)

or 4u – 4v = -1 ….(2)

Adding equation (1) and (2),

Subtituting this value of x in (3)

3 (1) + y = 4

or 3 + y = 4

or y = 4 – 3 = 1

Hence, = 1 and y = 1

Question 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.

Solution:

(i) Let the speed of Ritu in still water = x km/hr

and speed of the current = y km/hr

∴ Speed against the current = (x – y) km/hr and speed in the direction of the current = (x + y) km/hr

Distance covered by Ritu in the direction of the current in 2 hours

= Speed × Time

= (x + y) × 2 km

According to the first condition,

2 (x + y) = 20

x + y = 10 …(1)

Distance covered by Ritu against the current in 2 hours

= Distance × Time

= 2 (x – y) km

According to the second condition,

⇒ 2 (x – y)= 4

⇒ x – y= 2 …(2)

Adding equation (1) and (2),

Substituting this value of x in (1)

6 + y = 10

⇒ y = 10 – 6 = 4

Hence, Speed of Ritu in still water = 6 km/hr

andspeed of the current = 4 km/hr

(ii) Suppose that a woman can finish the work in = x days

a man can finish the work in = y days

Then, work of a woman of 1 day = \(\frac{1}{x}\)

and work of a man of 1 day = \(\frac{1}{y}\)

According to the first condition,

\(\frac{2}{x}\) + \(\frac{5}{y}\) = \(\frac{1}{4}\) …(1)

According to the second condition,

\(\frac{3}{x}\) + \(\frac{6}{y}\) = \(\frac{1}{3}\) …(2)

Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v. from (1) and (2)

2u + 5v = \(\frac{1}{4}\)

and 3u + 6v = \(\frac{1}{3}\)

or 8u + 20v = 1 …(3)

or 9u + 18v = 1 ….(4)

Multiplying equation (3) by 9 and equation (4) by 8

72u + 180v = 9 (5) …(5)

and 72u + 144v = 8 …(6)

Subtracting (6) from equation (5),

Hence, a woman and a man can complete the work alone in 18 days and 36 days respectively.

(iii) Let the speed of thc train = x km/br

and the speed of the bus= y km/hr

Total distance 300 km

Case I

Time taken in covering a distance 60 km

= \(\frac{Distance}{Speed}\)

= \(\frac{60}{x}\)

Time taken by bus in covering a distance (300 – 60 = 240) km.

= \(\frac{240}{y}\) hours

∴ Total time = \(\left(\frac{60}{x}+\frac{240}{y}\right)\) hours

According to the first condition,

\(\frac{60}{x}\) + \(\frac{240}{y}\) = 4

⇒ \(\frac{15}{x}\) + \(\frac{60}{y}\) = 1 …(1)

Case II

Time taken by train in covering a distance 100 km

= \(\frac{100}{x}\)hours

Time taken by bus in covering a distance (300 – 100 = 200) km

= \(\frac{200}{y}\)hours

∴ Total time = \(\left(\frac{100}{x}+\frac{200}{y}\right)\) hours

According to the second condition,

Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v in equations (1) and (2)

15u + 60v = 1

and 24u + 48v = 1

or 15u + 60v – 1 = 0

24u + 48v – 1 = 0

Hence, the speeds of the train and the bus are respectively 60 km/hr and 80 km/hr.

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