Ex-3.3 Pair Of Linear Equations In Two Variables (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 17. 15/u + 2/v = 1/7

1/u + 1/v = 36/5

Sol. Let 1/x  = u

Let 1/y  = v

15x + 2y = 17 …………………………..(i)

x + y = 36/5……………………….(ii)

From equation (i) we get ,

2y = 17 − 15x

Substituting y =  in equation (ii) we get,

= 5( − 13x + 17) = 72

= − 65x = − 13

= x = 1/5

Putting x = 1/5 in equation (ii) , we get

1/5  + y = 36/5

= y = 7

= v = 1/y = 1/7

The Solution of the given system of equation is 5 and 1/7 respectively.

Question 18. 3/x − 1/y = − 9

2/x + 1/y = 5

Sol. Let 1/x  = u

Let 1/y  = v

3u − v = − 9…………………..(i)

2u + 3v = 5 ……………………….(ii)

Multiplying equation (i) *3 and (ii) *1 we get,

9u − 3v = − 27 ………………………….. (iii)

2u + 3v = 5 ……………………………… (iv)

Adding equation (i) and equation (iv) we get ,

9u + 2u − 3v + 3v = − 27 + 5

= u = − 2

Putting u = − 2 in equation (iv) we get,

2( − 2) + 3v = 5

= 3v = 9

= v = 3

Hence x = 1/u = − 1/2

Hence y = 1/v = 1/3

Question 19. 2/x − 3/y = 9/xy

2/x + 1/y = 9/xy

Sol. 2/x − 3/y = 9/xy ……………………… (i)

2/x + 1/y = 9/xy…………………….. (ii)

Multiplying equation (i) adding equation (ii) we get,

2y + 3x = 9………………………..(iii)

4y + 9x = 21 ……………………….(iv)

From (iii) we get ,

3x = 9 − 2y

Substituting x =  in equation (iv) we get

= 4y + 3(9 − 2y) = 21

= − 2y = 21 − 27

= y = 3

Putting y = 3 in x =  we get,

= x = 1

Hence the Solutions of the system of equation are 1 and 3 respectively.

Question 20. 1/5x + 1/6y = 1/2

1/5x + 1/6y = 8

Sol. Let 1/x  = u

Let 1/y  = v

u/5 + v/6 = 12

= 6u + 5v = 360 …………….(i)

u/3 + 3v/7 = 8

= 7u − 9v = 168 …………….(ii)

Let us eliminate v from the equation (i) and (ii) . multiplying equation (i) by 9 and (ii) by 5

54u + 35u = 3240 + 840

89u = 4080

= u = 4080/89

Putting u = 4080/89 in equation (i) we get,

6(4080/89) + 5v = 360

= 24480/89 + 5v = 360

= v = 7560/89

= v = 7560/5∗89

= v = 1512/89

1/u  = x = 89/4080

1/v  = y = 89/1512

.

Question 27.

Then, the given system of equation becomes,

6u = 7v + 3

6u − 7v = 3……………………….. (i)

And u/2 = v/3

3u = 2v

3u − 2v = 0 ……………………… (ii)

Multiplying equation (ii) by 2 and (i) 1

6u − 7v = 3

6u − 4v = 0

Subtracting v = − 1 in equation (ii) ,we get

3u − 2( − 1) = 0

3u + 2 = 0

3u = − 2

= u = − 2/3

x + y = − 3/2 …………………….(v)

and v = − 1

x − y = − 1……………………(vi)

Adding equation (v) and equation (vi) we get,

2x = − 3/2 − 1

= x = − 5/4

Putting x = − 2/3 in equation (vi)

= − 5/4 − y = − 1

= y = − 1/4

Question 28

Sol.

5xy = 6(x + y)

= 5xy = 6x + 6y ……………….(i)

And

xy = 6(y − x)

= xy = 6y − 6x ……………………(ii)

Adding equation (i) and equation (ii) we get,

6xy = 6y + 6y

6xy = 12y

x = 2

Putting x = 2 in equation (i) we get,

10y = 12 + 6y

10 − 6y = 12

4y = 12

y = 3

The Solution of the given system of equation is 2 and 3 respectively.

Question 29

Sol. Then the given system of equation becomes:

22u + 45v = 5 ……………………….. (i)

55u + 45v = 14 ………………………(ii)

Multiplying equation (i)by 3 and (ii) by 1

66u + 45v = 15 ……………………… (iii)

55u + 45v = 14 ……………………… (iv)

Subtracting equation (iv) from equation (iii) , we get

66u − 55u = 15 − 14

= 11u = 1

= u = 1/11

Putting = u = 1/11 in equation (i) we get,

= 2 + 15v = 5

= 15v = 3

= v = 15

Now,

= x + y = 11 ……………..(v)

1/x − y = v

= x − y = 5 …………………..(vi)

Adding equation (v) and (vi) we get,

2x = 16

= x = 8

Putting the value of x in equation (v)

8 + y = 11

= y = 3

The Solutions of the given system of equation are 8 and 3 respectively.

Question 30

Sol.

Then the given system of equation becomes:

5u − 2v = − 1 ……………………….. (i)

15u + 7v = 10 ………………………(ii)

Multiplying equation (i) by 7 and (ii) by 2

35u − 14v = − 7 ……………………… (iii)

30u + 14v = 20 ……………………… (iv)

Subtracting equation (iv) from equation (iii) , we get

− 2v = − 1 − 1

= − 2v = − 2

= v = 1

Now,

= x + y = 5 ……………..(v)

= x − y = 1 …………………..(vi)

Adding equation (v) and (vi) we get,

2x = 6

= x = 3

Putting the value of x in equation (v)

3 + y = 5

= y = 2

The Solutions of the given system of equation are 3 and 2 respectively.

Question 31. 

Sol.Then the given system of equation becomes:

3u + 2v = 2 ……………………….. (i)

9u + 4v = 1 ………………………(ii)

Multiplying equation (i) by 3 and (ii) by 1

6u + 4v = 4 ……………………… (iii)

9u − 4v = 1 ……………………… (iv)

Adding equation (iii) and (iv) we get,

45u = 5

= u = 3

Subtracting equation (iv) from equation (iii) , we get

2v = 2 − 1

= 2v = 1

= v = 1/2

Now,

= x + y = 3 ……………..(v)

= x − y = 2 …………………..(vi)

Adding equation (v) and (vi) we get,

2x = 5

= x = 5/2

Putting the value of x in equation (v)

5/2 + y = 11

= y = 1/2

The Solutions of the given system of equation are 5/2  and 1/2  respectively.

Question 32

Sol. Then the given system of equation becomes:

3u + 10v = − 9 ………………………..(i)

5u/4 − 3v/5 = 61/60

25u − 12v = 61/3 ………………………(ii)

Multiplying equation (i) by 12 and (ii) by 10

36u + 120v = − 108 ……………………… (iii)

250u + 120v = 610/3 ……………………… (iv)

Adding equation (iv) and equation (iii) , we get

36u + 250u = 610/3 − 108

= 286u = 286/3

= u = 1/3

Putting u = 61/3 in equation (i)

3(1/3) + 10v = − 9

= v = − 1

Now,

= x + 2y = 3 ……………..(v)

= 3x − 2y = − 1 …………………..(vi)

Putting x = 1/2 in equation (v) we get,

1/2 + 2y = 3

= y = 5/4

The Solutions of the given system of equation are 1/2

And 5/4 respectively.