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

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

Question 1.

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

Solution:

1. The given pair of equations is

Taking \(\frac{1}{x}=a\) and \(\frac{1}{y}=b\) in both the equations, we get

\(\frac{1}{2}\)a + \(\frac{1}{3}\)b = 2 ……….(3)

\(\frac{1}{3}\)a + \(\frac{1}{2}\)b = \(\frac{13}{6}\) ………..(4)

Multiplying both the equations by 6, we get

3a + 2b = 12 ……(5)

2a + 3b = 13 ……(6)

Adding equations (5) and (6), we get

5a + 5b = 25

∴ a + b = 5 ……….(7)

Subtracting equation (6) from equation (5), we get

a – b = -1 ………..(8)

Equations (7) and (8) can be solved easily to get a = 2 and b = 3.

Then a = \(\frac{1}{x}\) = 2 and b = \(\frac{1}{y}\) = 3

∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\)

Thus, the solution of the given pair of equations is x = \(\frac{1}{2}\), y = \(\frac{1}{3}\)

2. The given pair of equations is

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

\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\) ……….(2)

Taking \(\frac{1}{\sqrt{x}}=a\) and \(\frac{1}{\sqrt{y}}=b\) in both the equations, we get

2a + 3b = 2 ……….(3)

4a – 9b = -1 ……….(4)

Multiplying equation (3) by 3, we get

6a + 9b = 6 ……….(5)

Adding equations (4) and (5). we get

(4a – 9b) + (6a + 9b) = -1 + 6

∴ 10a = 5

∴ a = \(\frac{1}{2}\)

Substituting a = \(\frac{1}{2}\) in equation (3), we get

2(\(\frac{1}{2}\)) + 3b = 2

∴ 1 + 3b = 2

∴ 3b = 1

∴ b = \(\frac{1}{3}\)

Now, a = \(\frac{1}{\sqrt{x}}=\frac{1}{2}\)

∴ 2 = \(\sqrt{x}\)

∴ x = 4

Again b = \(\frac{1}{\sqrt{y}}=\frac{1}{3}\)

∴ 3 = \(\sqrt{y}\)

∴ y = 9

Thus, the solution of the given pair of equations is x = 4, y = 9,

3. The given pair of equations is

\(\frac{4}{x}\) + 3y = 14 ……….(1)

\(\frac{3}{x}\) – 4y = 23 ……….(2)

Taking \(\frac{1}{x}\) = a in both the equations, we get

4a + 3y = 14 ……….(3)

3a – 4y = 23 ……….(4)

Multiplying equation (3) by 4 and equation (4) by 3 and adding them, we get

4(4a + 3y) + 3(3a – 4y) = 4(14) + 3(23)

∴ 16a + 12y + 9a – 12y = 56 + 69

∴ 25a = 125

∴ a = 5

Now, a = \(\frac{1}{x}\) = 5

∴ x = \(\frac{1}{5}\)

Substituting x = \(\frac{1}{5}\) in equation (1), we get

\(\frac{4}{\frac{1}{5}}\) + 3y = 14

∴ 20 + 3y = 14

∴ 3y = -6

∴ y = -2

Thus, the solution of the given pair of equations is x = \(\frac{1}{5}\), y = -2.

4. The given pair of equations is

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

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

Taking \(\frac{1}{x-1}=a\) and \(\frac{1}{y-2}=b\) in both the equations, we get

5a + b = 2 ………..(3)

6a – 3b = 1 ………..(4)

Multiplying equation (3) by 3 and then additing equation (4) to it, we get

3(5a + b) + (6a – 3b) = 3(2) + 1

∴ 15a + 3b + 6a – 3b = 6 + 1

∴ 21a = 7

a = \(\frac{1}{3}\)

Substituting a = \(\frac{1}{3}\) in equation (4), we get

6(\(\frac{1}{3}\)) – 3b = -1

∴ 2 – 3b = 1

∴ 1 = 3b

∴ b = \(\frac{1}{3}\)

Now, a = \(\frac{1}{x-1}=\frac{1}{3}\)

∴ x – 1 = 3

∴ x = 4

Again, b = \(\frac{1}{y-2}=\frac{1}{3}\)

∴ y – 2 = 3

∴ y = 5

Thus, the solution of the given pair of equations is x = 4, y = 5.

5. The given pair of equations is \(\frac{7 x-2 y}{x y}=5\) and \(\frac{8 x+7 y}{x y}=15\)

Hence, \(\frac{7}{y}-\frac{2}{x}=5\) ………..(1)

and \(\frac{8}{y}+\frac{7}{x}=15\) ………..(2)

Taking \(\frac{1}{y}\) = a and \(\frac{1}{y}\) = b in both the equations, we get

7a – 2b = 5 ………..(3)

8a + 7b = 15 ………..(4)

Expressing the equation in the standard form, we get

7a – 2b – 5 = 0 and 8a + 7b – 15 = 0

Now, a = \(\frac{1}{y}\) = 1 ∴ y = 1

and b = \(\frac{1}{x}\) = 1 ∴ x = 1

Thus, the solution of the given pair of equations is x = 1, y = 1.

6. The given pair of equations is 6x + 3y = 6xy and 2x + 4y = 5xy

Dividing both the equations by xy, we get

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

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

Taking \(\frac{1}{y}\) = a and \(\frac{1}{x}\) = b,

we get

6a + 3b = 6

i.e., 2a + b = 2 ……(3)

2a + 4b = 5 ……(4)

Subtracting equation (3) from equation (4),

we get

(2a + 4b) – (2a + b) = 5 – 2

∴ 3b = 3

∴ b = 1

Substituting b = 1 in equation (3), we get

2a + 1 = 2

∴ 2a = 1

∴ a = \(\frac{1}{2}\)

Now, a = \(\frac{1}{y}=\frac{1}{2}\) ∴ y = 2

and b = \(\frac{1}{x}\) = 1 ∴ x = 1

Thus, the solution of the given pair of equations is x = 1, y = 2.

7. The given pair of equations is

\(\frac{10}{x+y}+\frac{2}{x-y}=4\) ………..(1)

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

Taking \(\frac{1}{x+y}=a\) and \(\frac{1}{x-y}=b\) in both the equations, we get

10a + 2b = 4

i.e., 5a + b = 2 ……(3)

15a – 5b = -2 ……(4)

Multiplying equation (3) by 5 and then adding equation (4) to it, we get

5(5a + b) + (15a – 5b) = 5(2) + (-2)

∴ 25a + 5b + 15a – 5b = 10 – 2

∴ 40a = 8

∴ a = \(\frac{1}{5}\)

Substituting a = \(\frac{1}{5}\) in equation (3), we get

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

∴ 1 + b = 2

∴ b = 1

Now, a = \(a=\frac{1}{x+y}=\frac{1}{5}\)

∴ x + y = 5 ……(5)

and b = \(\frac{1}{x-y}=1\)

∴ x – y = 1 ………(6)

Adding equations (5) and (6), we get

2x = 6

∴ x = 3

Substituting x = 3 in equations (5), we get

3 + y = 5

∴ y = 2

Thus, the solution of the given pair of equations is x = 3, y = 2.

8. The given pair of equations is

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

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

Taking \(\frac{1}{3 x+y}=a\) and \(\frac{1}{3 x-y}=b\) in both the equations, we get

a + b = \(\frac{3}{4}\) ………(3)

\(\frac{a}{2}-\frac{b}{2}=-\frac{1}{8}\)

i.e., a – b = –\(\frac{2}{8}\)

i.e., a – b = –\(\frac{1}{4}\) ………….(4)

Adding equations (3) and (4), we get

2a = \(\frac{2}{4}\)

∴ a = \(\frac{1}{4}\)

Substituting a = \(\frac{1}{4}\) in equation (3), we get

\(\frac{1}{4}+b=\frac{3}{4}\)

∴ b = \(\frac{1}{2}\)

Now, a = \(\frac{1}{3 x+y}=\frac{1}{4}\)

3x + y = 4 ………….(5)

and b = \(\frac{1}{3 x-y}=\frac{1}{2}\)

∴ 3x – y = 2 ………….(6)

Adding equations (5) and (6), we get

6x = 6

∴ x = 1

Substituting x = 1 in equations (5), we get

3(1) + y = 4

∴ y = 1

Thus, the solution of the given pair of equations is x = 1, y = 1.

2. Formulate the following problems as a pair of equations, and hence find their solutions:

Question 1.

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.

Solution:

Let Ritu’s speed of rowing in still water be x km/h and the speed of the current by y km/h.

Then, her net speed going down-stream = (x + y) km/h and her net speed going upstream = (x – y) km/h.

Also, time = \(\frac{\text { distance }}{\text { speed }}\)

Then, form the first condition, we get

2 = \(\frac{20}{x+y}\)

∴ x + y = 10 ……(1)

And, from the second condition, we get

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

∴ x – y = 2 ………..(2)

Adding equations (1) and (2), we get

2x = 12

∴ x = 6

Substituting x = 6 in equation (1), we get

6 + y = 10

∴ y = 4

Thus, Ritu’s speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h.

Question 2.

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.

Solution:

Suppose that 1 woman alone can finish the work in x days and 1 man alone can finish the work in y days.

∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\) part and work done by 1 man in 1 day \(\frac{1}{y}\) part.

Then, work done by 2 women and 5 men together in 1 day = \(\left(\frac{2}{x}+\frac{5}{y}\right)\) part.

But, according to the first information given, 2 women and 5 men together finish the work in 4 days, i.e., in one day they can finish \(\frac{1}{4}\) part of the work.

Hence, we get the following equation:

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

Similarly, from the second information given, we get

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

Taking \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b in both the equations, we get

2a + 5b = \(\frac{1}{4}\) ……(3)

3a + 6b = \(\frac{1}{3}\) ……(4)

Multiplying equation (3) by 6 and equation (4) by 5, we get

12a + 30b = \(\frac{6}{4}\) ……(5)

15a + 30b = \(\frac{5}{3}\) ……(6)

Subtracting equation (5) from equation (6), we get

(15a + 30b) – (12a + 30b) = \(\frac{5}{3}-\frac{6}{4}\)

15a + 30b – 12a – 30b = \(\frac{20-18}{12}\)

∴ 3a = \(\frac{2}{12}\)

∴ a = \(\frac{1}{18}\)

Substituting a = \(\frac{1}{18}\) in equation (3), we get

2(\(\frac{1}{18}\)) + 5b = 4

∴ 5b = \(\frac{1}{4}-\frac{1}{9}\)

∴ 5b = \(\frac{5}{36}\)

∴ b = \(\frac{1}{36}\)

Now, a = \(\frac{1}{x}=\frac{1}{18}\) ∴ x = 18

and b = \(\frac{1}{y}=\frac{1}{36}\) ∴ y = 36

Thus, 1 woman alone can finish the work in 18 days and 1 man alone can finish the work in 36 days.

Question 3.

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:

Let the speed of the train be x km/h and the speed of the bus be y km/h.

Also, time = \(\frac{\text { distance }}{\text { speed }}\)

In the first case, she travels 60 km by train and the remaining 240 km (300 – 60) by bus.

∴ Time taken for journey by train = \(\frac{60}{x}\)h and time taken for journey by bus \(\frac{240}{y}\)h.

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

In the first case, total time taken = 4 h.

Hence, we get the following equation:

\(\frac{60}{x}+\frac{240}{y}=4\) ………..(1)

Similarly, in the second case, the distances she travels by train and bus are 100 km and 200 km respectively and time taken by those journies are \(\frac{100}{x}\)h and \(\frac{200}{y}\)h respectively.

In this case, the total time taken = 4 hours + 10 minutes = 4\(\frac{1}{6}\) hours.

Hence, we get the following equation:

\(\frac{100}{x}+\frac{200}{y}=4 \frac{1}{6}\)

∴ \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\) ………..(2)

Taking \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b in both the equations, we get

60a + 240b = 4 ………..(3)

100a + 200b = \(\frac{25}{6}\) ………..(4)

Multiplying equation (3) by 5 and equation (4) by 6, we get

300a + 1200b = 20 ………..(5)

600a + 1200b = 25 ………..(6)

Subtracting equation (5) from equation (6), we get

(600a + 1200b) – (300a + 1200b) = 25 – 20

∴ 300a = 5

∴ a = \(\frac{1}{60}\)

Substituting a = \(\frac{1}{60}\) in (3), we get

60(\(\frac{1}{60}\)) + 240b = 4

∴ 1 + 240b = 4

∴ 240b = 3

∴ b = \(\frac{1}{80}\)

Now, a = \(\frac{1}{x}\) = \(\frac{1}{60}\)

∴ x = 60

and b = \(\frac{1}{y}\) = \(\frac{1}{80}\)

∴ y = 80

Thus, the speed of the train is 60 km/h and the speed of the bus is 80 km/h.