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

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

Question 1.

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

Solution:

(i) The given pair of equations is

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

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

Put \(\frac { 1 }{ x } =a\) …………………. (3)

and \(\frac { 1 }{ y } =b\) ……………… (4)

Then the equations (1) and (2) can be rewritten as

\(\frac { 1 }{ 2 } a+\frac { 1 }{ 3 } b=2\) ………………… (5)

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

3a + 2b = 12 ………….. (7)

2a + 3b = 13 …………….. (8)

Multiplying equations (7) by 3 and equation (8) by 2, we get

9a + 6b = 36 …………… (9)

4a + 6b = 26 ………….. (10)

Subtracting equation (10) from equation (9), we get

5a = 10

a = \(\frac { 10 }{ 5 } =2\) ………………. (11)

Substituting this”value of a in equation (9), we get

9(2) + 6b = 36

18 + 6b = 36

6b = 36 – 18 = 18

b = \(\frac { 18 }{ 6 }\) = 3 …………….. (12)

From equation (3) and equation (11), we get

\(\frac { 1 }{ x }\) = 2

x = \(\frac { 1 }{ 2 }\)

From equation (4) and equation (11), we get

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

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

Hence, the solution of the given pair of equations is

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

Verification: Substituting x = \(\frac { 1 }{ 2 }\), y = \(\frac { 1 }{ 3 }\), we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution we have got is correct.

(ii) The given pair of equations is

Then equations (1) and (2) can be rewritten as

2u + 3υ = 2 ……………. (5) υ u

4u – 9υ = – 1 …………… (6)

MultipLying equation (5) by 3, we get

6u + 9υ = 6 …………….(7)

Adding equation (6) and equation (7), we get

10u = 5

u = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\) ………… (8)

Substituting the value of u in equation (5), we get

\(\left( \frac { 1 }{ 2 } \right)\) = 3υ = 2

1 + 3υ = 2

3υ = 2 – 1 = 1 …………… (9)

υ = \(\frac { 1 }{ 3 }\)

From equation (3) and equation (8), we get

\(\frac { 1 }{ \sqrt { x } } \) = \(\frac { 1 }{ 2 }\)

\(\sqrt { x }\) = 2

x = 4

From equation (4) and equation (9), we get

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

\(\sqrt { y }\) = 3

y = 9

Hence, the solution of the given pair of equation is x = 4, y = 9.

Verification: Substituting x = 4, y = 9,

we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution we have got is correct.

(iii) \(\frac { 4 }{ x }\) + 3y = 14 ………….. (1)

\(\frac { 3 }{ x }\) – 4y = 23 …………….. (2)

Put \(\frac { 1 }{ x }\) = X

The equation (1) and (2) can be written as

4X + 3y = 14 ……………..(4)

3X – 4y = 23 ………………(5)

From equation (5),

4y = 3X – 23

\(y=\frac { 3X-23 }{ 4 } \)

Substitute this value ofy in equation (4), we get

\(4X+3\left( \frac { 3X-23 }{ 4 } \right) =14\)

16X + 9X – 69 = 56

25X = 56 + 69 = 125

X = \(\frac { 125 }{ 25 }\) = 5 ………….. (7)

Substituting this value of X in equation (6), we get

\(y=\frac { 3(5)-23 }{ 4 } =\frac { 15-23 }{ 4 } \)

= \(\frac { – 8 }{ 4 }\) = – 2 ……………….. (8)

From equation (3) and equation (7), we get

\(\frac { 1 }{ x }\) = 5

x = \(\frac { 1 }{ 5 }\) ……….(9)

Hence, the solution of the given pair of

equations is x = \(\frac { 1 }{ 5 }\), y = – 2.

Verification: Substituting x = \(\frac { 1 }{ 5 }\), y = – 2,

we find that both the equations (1) and (2)

are satisfied as shown below:

Hence, the solution is correct.

(iv) The given pair of equations is

Then equations (1) and (2) can be written as

5u + υ = 2 ……………..(5)

6u – 3υ = 1 ……………(6)

Multiplying equation (5) by 3, we get

15u + 3υ = 6

Adding equation (6) and equation (7), we get

21u + 7

u = \(\frac { 7 }{ 21 }\) = \(\frac { 1 }{ 3 }\) ………….. (8)

Substituting this value of u in equation (5), we get

From equation (3) and equation (s), we get

\(\frac { 1 }{ x-1 } \) = \(\frac { 1 }{ 3 }\)

x – 1 = 3

x = 1+ 3 = 4

From equation (4) and equation (9), we get

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

y – 2 = 3

y = 3 + 2 = 5

Hence, the solution of the given pair of

equations is x = 4, y = 5.

Verification: Substituting x = 4, y = 5, we

find that both the equations (1) and (2) are

satisfied as shown below:

Hence, the solution is correct.

(v) The given pair of equations is

Then the equations (3) and (4) can be rewritten as

7υ – 2u = 5

8υ + 7u = 15

7υ – 2u – 5 = 0

8υ + 7u – 15 = 0

To solve the equations by the cross multiplication method, we draw the diagram below:

From equation (5) and equation (10), we get

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

x = 1

From equation (6) and equation (9), we get

\(\frac { 1 }{ y }\) = 1

y = 1

Hence, the solution of the given pair of

equations is x = 1, y = 1.

Verification: Substituting x = 1, y = 1 we

find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

(vi) The given pair of equations is

6x + 3y = 6xy

2x + 4y = 5xy

(Dividing throughout by xy)

Then equations (1) and (2) can be rewritten as

6υ + 3u = 6

2υ + 4u = 5

Multiplying equation (6) by 3, we get

6υ + 12u = 15

Subtracting equation (5) from equation (7), we get

9u = 9

u = \(\frac { 9 }{ 9 }\) = 1

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

x = 1 [ using (3)]

Substituting this value of u in equation (5), we get

6u + 3 x 1 = 6

6υ + 3 = 6

6υ = 6 – 3 = 3

υ = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

\(\frac { 1 }{ y }\) = \(\frac { 1 }{ 2 }\)

y = 2 [ using (4)]

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

Verification: Substituting x = 1, y = 2, we

find that both the equations (1) and (2) are satisfied as shown below:

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

= 3 + 3 = 6

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

= 1+ 4 = 5

Hence, the solution is correct.

(vii) The given pair of equations is

Then equations (1) and (2) can be written as

10u + 2υ = 4 ………….(5)

15u – 5υ = – 2 ………….. (6)

Dividing (5) throughout by 2, we have

5u + υ = 2 ………… (7)

Multiplying equation (7) by 3, we get

15u + 3υ = 6 ………..(8)

Subtracting equation (6) from equation (8), we get

8υ = 8

υ = \(\frac { 8 }{ 8 }\) = 1 …………….(9)

Substituting this value of u in equation (7), we get

Su + 1=2

Substituting this value of u in equation (7), we get

5u + 1 = 2

5u = 2 – 1 = 1

u = \(\frac { 1 }{ 5 }\) ………………..(10)

From equation (3) and equation (10), we get

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

= x + y = 5 ………………(11)

From equation (4) and equation (9), we get

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

x – y = 1 ……………… (12)

Adding equation (11) and equation (12), we get

2x = 6

x = \(\frac { 6 }{ 2 }\) = 3

Substitute this value of x in equation (11), we get

3 + y = 5

y = 5 – 3 = 2

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

Verification: Substituting x = 3, y = 2,

we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

(viii) The given pair of equations is

Then equation (1) and (2) can be rewritten as

u + υ = \(\frac { 3 }{ 4 }\) ……………. (5)

\(\frac { 1 }{ 2 }\) u – \(\frac { 1 }{ 2 }\) υ = – \(\frac { 1 }{ 8 }\) ………….. (6)

(Multiplying both sides of(6) by 2) gives

u – υ = – \(\frac { 1 }{ 4 }\) …………… (7)

Adding equation (5) and equation (7), we get

2u = \(\frac { 3 }{ 4 }\) – \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

u = \(\frac { 1 }{ 4 }\) …………… (8)

Subtracting equation (7) from equation (5), we get

2υ = \(\frac { 3 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = 1

υ = \(\frac { 1 }{ 2 }\) …………… (9)

From equation (3) and equation (8), we get

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

3x + y = 4

From equation (4) and equaation (9), We gete

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

3x – y = 2

Adding equation (10) and equation (11), we get

6x = 6

x = \(\frac { 6 }{ 6 }\) = 1

Substituting this value of x in equation (10), we get

3(1) + y = 4

3 + y = 4

y = 4 – 3 = 1

Hence, the solution of the given pair of

equations is x = 1, y = 1.

Verification: Substituting x = 1, y = 1,

we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.

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 her speed of rowing in still water be x kim/hour and the speed of the current be y km/hour.

Then, her speed of rowing downstream

= (x + y) km/hour

and, her speed of rowing upstream

= (x – y) km/hour

Also,

In the first case, when she goes 20 km

downstream, the time taken is 2 hours.

∴\(\frac { 20 }{ x+y }\)

= x + y = 10

In the second case, when she goes 4 km

upstream, the time taken is 2 hours.

∴\(\frac { 4 }{ x-y }\)

= x – y = 2 …………….(2)

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

2x = 12

x = \(\frac { 12 }{ 2 }\) = 6

Substituting this value of x in equation (1), we get

6 + y = 10

y = 10 – 6 = 4

Hence, the speed of her rowing in still water

is 6 km/hour and the speed of the current is 4 km/hour.

Verification: Substituting x = 6, y = 4, we

find that both the equations (1) and (2) are

satisfied as shown below:

x + y = 6 + 4 = 10

x – y = 6 – 4 = 2

Hence, the solution is correct.

(ii) Let the time taken by 1 woman alone to finish the embroidery work be x days and the time taken by 1 man alone to finish the embroidery be y days.

Then

1 woman’s 1 day’s work = \(\frac { 1 }{ x }\)

and 1 man’s 1 day’s work = \(\frac { 1 }{ y }\)

∴ 2 womens 1 day’s work= \(\frac { 2 }{ x }\)

and 5 men’s 1 day’s work = \(\frac { 5 }{ y }\)

2 women and 5 men can ‘togethcr finish one embroidery work in 4 days.

∴\(4=\left( \frac { 2 }{ x } +\frac { 5 }{ y } \right) =1\)

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

Again, 3 women’s 1 day’s work = \(\frac { 3 }{ x }\)

and 6 men’s 1 day’s work = \(\frac { 6 }{ y }\)

3 women and 6 men can together finish one embroidery work in 3 days.

\(3=\left( \frac { 3 }{ x } +\frac { 6 }{ y } \right) =1\)

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

Put \(\frac { 1 }{ x }\) = u …………… (3)

Put \(\frac { 1 }{ y }\) = υ ………………….. (4)

Then equations (1) and (2) can be rewritten as

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

3u + 6υ = \(\frac { 1 }{ 3 }\) …………….. (6)

Multiplying equation (5) by 3 and equation (6) by 2, we get

6u + 15υ = \(\frac { 3 }{ 4 }\) …………… (7)

6u + 12υ = \(\frac { 2 }{ 3 }\) ………….. (8)

Subtracting equation (8) from equation (7), we get

3υ = \(\frac { 3 }{ 4 }\) – \(\frac { 2 }{ 3 }\) = \(\frac { 1 }{ 12 }\)

υ = \(\frac { 1 }{ 36 }\) ………………… (9)

Substituting this value of y in equation (5), we get

2u + \(5\left( \frac { 1 }{ 36 } \right)\) = \(\frac { 1 }{ 4 }\)

2u + \(\frac { 5 }{ 36 }\) = \(\frac { 1 }{ 4 }\)

2u = \(\frac { 1 }{ 4 }\) – \(\frac { 5 }{ 36 }\)

2u = \(\frac { 9 }{ 36 }\) – \(\frac { 5 }{ 36 }\)

2u = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)

u = \(\frac { 1 }{ 18 }\) ………….. (10)

From equation (3) and equation (10), we get

\(\frac { 1 }{ x }\) = \(\frac { 1 }{ 18 }\)

x = 18

From equation (4) and equation (9), we get

\(\frac { 1 }{ y }\) = \(\frac { 1 }{ 36 }\)

y = 36

Hence, the time taken by 1 woman alone to finish the embroidery work is 18 days and the time taken by 1 man alone to finish the embroidery work is 36 days.

Verification : Substituting x = 18, y = 36,

we find that both the equations (1) and (2) are satisfied as shown below:

(iii) Let the speed of the train and the bus be x km/hour and y km/hour respectively.

When she travels 60 km by train and the remaining (300 – 60) km, i.e., 240 km by bus, the time taken is 4 hours.

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

Dividing throughout by 60,

\(\frac { 1 }{ x }\) + \(\frac { 4 }{ y }\) = \(\frac { 1 }{ 15 }\)

When she travels loo km by train and the remaining (300 – 100) km, i.e., 200 km by bus, the time taken is 4 hours 10 minutes, i.e., \(\frac { 25 }{ 6 }\) hours.

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

Dividing throughout by 25,

\(\frac { 4 }{ x }\) + \(\frac { 8 }{ y }\) = \(\frac { 1 }{ 6 }\) ……………… (2)

Multiplying (1) by 2, we get

\(\frac { 2 }{ x }\) + \(\frac { 8 }{ y }\) = \(\frac { 1 }{ 6 }\) ………….. (3)

Subtracting equation (3) from equation (2), we get

\(\frac { 2 }{ x }\) = \(\frac { 1 }{ 6 }\) – \(\frac { 2 }{ 15 }\) = \(\frac { 1 }{ 30 }\)

x = 60

Substituting this value of x in equation (3), we get

\(\frac { 2 }{ 60 }\) + \(\frac { 8 }{ y }\) = \(\frac { 2 }{ 15 }\)

\(\frac { 1 }{ 30}\) + \(\frac { 8 }{ y }\) = \(\frac { 2 }{ 15 }\)

\(\frac { 8 }{ y }\) = \(\frac { 2 }{ 15 }\) – \(\frac { 1 }{ 30 }\) = \(\frac { 1 }{ 10 }\)

y = 80

So, the solution of the equations (1) and (2)

is x = 60 and y = 80.

Hence, the speed of the train is 60 km/hour and the speed of the bus is 80 km/hour.

Verification: Substituting x= 60, y = 80,

we find that both the equations (1) and (2) are satisfied as shown below:

Hence, the solution is correct.