﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 3.6

Question-1 :-  Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2 and 1/3x + 1/2y = 13/6
(ii) 2/√x + 3/√y = 2 and 4/√x - 9/√y = -1
(iii) 4/x + 3y = 14 and 3/x – 4y = 23
(iv) 5/(x - 1) + 1/(y – 2) = 2 and 6/(x - 1) – 3/(y – 2) = 1

Solution :-
```(i) 1/2x + 1/3y = 2   .......(i)
1/3x + 1/2y = 13/6  ......(ii)

Let 1/x = a and 1/y = b
Then,
a/2 + b/3 = 2
3a + 2b = 6 .......(iii)

a/3 + b/2 = 13/6
2a + 3b = 13 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
3a + 2b = 12 ........X 2
2a + 3b = 13 .......X 3

Now,
6a + 4b = 24 ....(iii)
6a + 9b = 39 .....(iv)

By substracting eq (iv) and (iii)
6a + 4b - 6a - 9b = 24 - 39
-5b = -15
b = 15/5
b = 3

Put the value of b = 3 in Equation (iv)
2a + 3 x 3 = 13
2a = 13 - 9
2a = 4
a = 4/2
a = 2

Hence, x = 1/a = 1/2 and y = 1/b = 1/3
```
```(ii) 2/√x + 3/√y = 2  .......(i)
4/√x - 9/√y = -1  ......(ii)

Let 1/√x = a and 1/√y = b
Then,
2a + 3b = 2 .......(iii)
4a - 9b = -1 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
2a + 3b = 2 ........X 4
4a - 9b = -1 .......X 2

Now,
8a + 12b = 8 ....(iii)
8a - 18b = -2 .....(iv)

By substracting eq (iv) and (iii)
8a + 12b - 8a + 18b = 8 + 2
30b = 10
b = 10/30
b = 1/3

Put the value of b = 1/3 in Equation (iv)
2a + 3 x 1/3 = 2
2a + 1 = 2
2a = 2 - 1
a = 1/2

Hence, √x = 1/a = 2 and √y = 1/b = 3
Then, x = 4, y = 9
```
```(iii) 4/x + 3y = 14  .......(i)
3/x – 4y = 23  ......(ii)

Let 1/x = a
Then,
4a + 3y = 14 .......(iii)
3a - 4y = 23 .......(iv)

For equal of any one variable(a or y), mutiplied by their coefficients following below:
4a + 3y = 14 .......X 3
3a - 4y = 23 .......X 4

Now,
12a + 9y = 42 ....(iii)
12a - 16y = 92 .....(iv)

By substracting eq (iv) and (iii)
12a + 9y - 12a + 16y = 42 - 92
25y = -50
y = -50/25
y = -2

Put the value of y = -2 in Equation (iv)
3a - 4 x (-2) = 23
3a = 23 - 8
3a = 15
a = 15/3
a = 5

Hence, x = 1/a = 1/5 and y = -2
```
```(iv) 5/(x - 1) + 1/(y – 2) = 2  .......(i)
6/(x - 1) – 3/(y – 2) = 1  ......(ii)

Let 1/(x - 1) = a and 1/(y - 2) = b
Then,
5a + b = 2 .......(iii)
6a - 3b = 1 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
5a + b = 2 ........X 6
6a - 3b = 1 .......X 5

Now,
30a + 6b = 12 ....(iii)
30a - 15b = 5 .....(iv)

By substracting eq (iv) and (iii)
30a + 6b - 30a + 15b = 12 - 5
21b = 7
b = 7/21
b = 1/3

Put the value of b = 1/3 in Equation (iv)
5a + 1/3 = 2
5a = 2 - 1/3
5a = 5/3
a = 1/3

Hence, 1/(x - 1) = a, 1/(y - 2)
x - 1 = 1/a, y - 2 = 1/b
x = 3 + 1, y = 3 + 2
x = 4, y = 5

```

(v) (7x - 2y)/xy = 5 and (8x + 7y)/xy = 15
(vi) 6x + 3y = 6xy and 2x + 4y = 5xy
(vii) 10/(x + y) + 2/(x - y) = 4 and 15/(x + y) – 5/(x - y) = -2
(viii) 1/(3x + y) + 1/(3x – y) = 3/4 and 1/2(3x + y) – 1/2(3x – y) = -1/8

Solution :-
```(v) (7x - 2y)/xy = 5
7/y - 2/x = 5 .......(i)

(8x + 7y)/xy = 15
8/y + 7/x = 15 ......(ii)

Let 1/x = a and 1/y = b
Then,
7b - 2a = 5 .......(iii)
8b + 7a = 15 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
7b - 2a = 5 ........X 8
8b + 7a = 15 .......X 7

Now,
56b - 16a = 40 ....(iii)
56b + 49a = 105 .....(iv)

By substracting eq (iv) and (iii)
56b - 16a - 56b - 49a = 40 - 105
-65a = -65
a = 65/65
a = 1

Put the value of a = 1 in Equation (iv)
8b + 7 x 1 = 15
8b = 15 - 7
8b = 8
b = 8/8
b = 1

Hence, x = 1/a = 1, y = 1/b = 1
```
```(vi) 6x + 3y = 6xy
2x + 4y = 5xy

Divide by xy both sides
6/y + 3/x = 6 .......(i)
2/y + 4/x = 5 ......(ii)

Let 1/x = a and 1/y = b
Then,
6b + 3a = 6 .......(iii)
2b + 4a = 5 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
6b + 3a = 6 .......X 2
2b + 4a = 5 .......X 6

Now,
12b + 6a = 12 ....(iii)
12b + 24a = 30 .....(iv)

By substracting eq (iv) and (iii)
12b + 6a - 12b - 24a = 12 - 30
-18a = -18
a = 18/18
a = 1

Put the value of a = 1 in Equation (iv)
2b + 4 x 1 = 5
2b = 5 - 4
2b = 1
b = 1/2
b = 1

Hence, x = 1/a = 1, y = 1/b = 2
```
```(vii) 10/(x + y) + 2/(x - y) = 4  .......(i)
15/(x + y) – 5/(x - y) = -2   ......(ii)

Let 1/(x + y) = a and 1/(x - y) = b
Then,
10a + 2b = 4 .......(iii)
15a - 5b = -2 .......(iv)

For equal of any one variable(a or y), mutiplied by their coefficients following below:
10a + 2b = 4 .......X 15
15a - 5b = -2 ......X 10

Now,
150a + 30b = 60 ....(iii)
150a - 50y = -20 .....(iv)

By substracting eq (iv) and (iii)
150a + 30b - 150a + 50b = 60 + 20
80b = 80
b = 80/80
b = 1

Put the value of b = 1 in Equation (iii)
10a + 2 x 1 = 4
10a = 4 - 2
10a = 2
a = 2/10
a = 1/5

Hence, 1/(x + y) = a = 1/5
x + y = 5 .....(v)

And 1/(x - y) = b = 1
x - y = 1 .....(vi)

Let us consider eq (vi)
x - y = 1
x = 1 + y ......(vii)

Put the value of x = 1 + y in eq (v)
1 + y + y = 5
2y = 5 - 1
2y = 4
y = 4/2
y = 2

Put the value of y = 2 in eq (vii)
x = 1 + 2
x = 3

x = 3, y = 2
```
```(viii) 1/(3x + y) + 1/(3x – y) = 3/4 .......(i)
1/2(3x + y) – 1/2(3x – y) = -1/8  ......(ii)

Let 1/(3x + y) = a and 1/(3x - y) = b
Then,
a + b = 3/4
4a + 4b = 3  .......(iii)

a/2 - b/2 = -1/8
a - b = -2/8
a - b = -1/4
4a - 4b = -1  .......(iv)

By adding eq (iv) and (iii)
4a + 4b + 4a - 4b = 3 - 1
8a = 2
a = 2/8
a = 1/4

Put the value of a = 1/4 in Equation (iii)
4 x 1/4 + 4b = 3
4b = 3 - 1
4b = 2
b = 2/4
b = 1/2

Hence, 1/(3x + y) = a = 1/4
3x + y = 4 .....(v)

And 1/(3x - y) = b = 1/2
3x - y = 2 .....(vi)

By adding eq (v) and eq (vi)
3x + y + 3x - y = 4 + 2
6x = 6
x = 6/6
x = 1

Put the value of x = 1 in eq (v)
3 x 1 + y = 4
3 + y = 4
y = 4 - 3
y = 1

x = 1, 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 still water be x km/h and speed of stream water be y km/h.
So, Upstream = x - y km/h
Downstream = x + y km/hs

2(x + y) = 20
x + y = 10  .......(i)

2(x - y) = 4
x - y = 2  ......(ii)

By adding eq (i) and (ii)
x + y + x - y = 10 + 2
2x = 12
x = 12/2
x = 6

Put the value of x = 6 in Equation (iv)
6 + y = 10
y = 10 - 6
y = 4

Hence, speed of Ritu in still water is 6 km/h and speed of stream water is 4 km/h.
```
```(ii) Let the no. of days taken by a woman x and by man y.
So, workdone by a woman = 1/x
workdone by a man = 1/y
According to question :

4(2/x + 5/y) = 1
8/x + 20/y = 1 .......(i)

3(3/x + 6/y) = 1
9/x + 18/y = 1  ......(ii)

Let 1/x = a and 1/y = b
Then,
8a + 20b = 1 .......(iii)
9a + 18b = 1 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
8a + 20b = 1 .......X 9
9a + 18b = 1 .......X 8

Now,
72a + 180b = 9 ....(iii)
72a + 144b = 8 .....(iv)

By substracting eq (iv) and (iii)
72a + 180b - 72a - 144b = 9 - 8
36b = 1
b = 1/36

Put the value of b = 1/36 in Equation (iv)
9a + 18 x 1/36 = 1
9a + 1/2 = 1
9a = 1 - 1/2
a = 1/18

Hence, x = 1/a = 18 and y = 1/b = 36
Then, x = 18, y = 36
No. of days taken by a woman = 18
No. of days taken by a man = 36
```
```(iii) Let the speed of train be x km/h and speed of bus be y km/h
According to question :

60/x + 240/y = 4
4(15/x + 60/y) = 4
15/x + 60/y = 1  .......(i)

100/x + 200/y = 25/6
25(4/x + 8/y) = 25/6
4/x + 8/y = 1/6
24/x + 48/y = 1  ......(ii)

Let 1/x = a and 1/y = b
Then,
15a + 60b = 1 .......(iii)
24a + 48b = 1 .......(iv)

For equal of any one variable(a or b), mutiplied by their coefficients following below:
15a + 60b = 1 .......X 24
24a + 48b = 1 .......X 15

Now,
360a + 1440y = 24 ....(iii)
360a + 720y = 15 .....(iv)

By substracting eq (iv) and (iii)
360a + 1440b - 360a - 720b = 24 - 15
720b = 9
b = 9/720
b = 1/80

Put the value of b = 1/80 in Equation (iii)
15a + 60 x 1/80 = 1
15a = 1 - 6/8
15a = 2/8
15a = 1/4
a = 1/60

Hence, x = 1/a = 60 and b = 1/y = 80
The speed of train = 60 km/h
The speed of bus = 80 km/h
```
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