Ex-3.3 Pair Of Linear Equations In Two Variables (Part - 3), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 34

x + y = 5xy

3x + 2y = 13xy

Sol.

The given system of equations is:

x + y = 5xy ………………….. (i)

3x + 2y = 13xy……………… (ii)

Multiplying equation (i) by 2 and equation (ii) 1 we get,

2x + + 2y = 10xy ………………… (iii)

3x + 2y = 13xy ……………………. (iv)

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

3x − 2x = 13xy − 10xy

= x = 3xy

=  x/3x = y

=  1/3 = y

Putting y =  1/3 = y in equation (i) we get,

= x + y = 5(x)( 1/3)

= x + x/3x =  5x/3

= 2x = 1

= x =  1/2 = y

Hence Solution of the given system of equation is 1/2 and 1/3

Question 35. x + y = xy

Sol. x + y = xy ………………………………….. (i)

………………….. (ii)

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

2x = 2xy + 6xy

= 2x = 6xy

= y = x + y = xy

= y = 1/4

Putting = y = 1/4 in equation (i) , we get,

= x + 1/4 = 2x(1/4)

= x = − 1/2

Hence the Solution of the given system of equation is = x = − 1/2

And y = 1/4 respectively.

Question 36. 2(3u − v) = 5uv

2(u + 3v) = 5uv

Solution. 2(3u − v) = 5uv

= 6u − 2v = 5uv …………………. (i)

2(u + 3v) = 5uv

2u + 6v = 5uv …………………….. (ii)

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

18u − 6v = 15uv …………………….. (iii)

2u + 6v = 5uv ………………………….. (iv)

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

18u + 2u = 15uv + 5uv

= v = 1

Putting v = 1 in equation (i) we get,

6u − 2 = 5u

= u = 2

Hence the Solution of the given system of Solution of equation is 2 and 1 respectively.

Question 37. 

Sol. Then the given system of equation becomes:

2u + 3v = − 17/5 ………………………..(i)

5u − v = 2 ……………………… (ii)

Multiplying equation (ii) by 3

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

15u − 2u = − 17/5 + 5

= 13u = 13/5

= u = 1/5

Putting u = 1/5 in equation (i)

5(1/5) + v = − 2

= v = 1

Now,

= 3x + 2y = 5 ……………..(iv)

= 3x − 2y = 1 …………………..(v)

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

6x = 6

= x = 1

Putting the value of x in equation (v) we get,

3 + 2y = 5

= y = 1

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

Question 38. 

Sol. Then the given system of equation becomes:

44u + 30v = 10 ……………………….. (i)

55u + 40v = 13 ………………………(ii)

Multiplying equation (i) by 4 and (ii) by 3

176u + 120v = 40 ……………………… (iii)

165u + 120v = 39 ……………………… (iv)

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

176 − 165u = 40 − 39

= u = 1/11

Putting the value of u in equation (i)

44(1/11) + 30v = 10

= 4 + 30v = 10

= 30v = 6

= x + y = 11……………..(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 40

Sol. Then the given system of equation becomes:

10p + 2q = 4 ……………………….. (i)

15p − 5q = − 2 ………………………(ii)

Multiplying equation (i) by 4 and (ii) by 3

176u + 120v = 40 ……………………… (iii)

165u + 120v = 39 ……………………… (iv)

Using cross multiplication method we get,

p =  1/5 and q = 1

x + y = 5…………………….. 3

x − y = 1 ……………………..4

Adding equation 3 and 4 we get,

x = 3

Substituting the value of x in equation 3 we get,

y = 2

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

Question 41

Sol. Then the given system of equation becomes:

 2 ………………………………..1

 1……………………………………. 2

Can be written as 5p + q = 2 ………………………………… 3

6p − 3q = 1 ………………………………………. 4

Equation 3 and 4 from a pair of linear equation in the general form. Now, we  can use any method to solve these equations.

We get p = 1/3

q =  1/3

Substituting the  for p , we have

x − 1 = 3

x = 4

y − 2 = 3

y = 5

The Solution of the required pair of equation is 4 and 5 respectively.

Question 42

Sol. 7/y − 2/x = 5…………………………..1

8/y + 7/x = 15……………………………2

Let 1/x = p

Let 1/y = q

The given equation s reduce to:

− 2p + 7q = 5

= − 2p + 7q − 5 = 0 ……………………………….. 3

7p + 8q = 15

= 7p + 8q − 15 = 0 …………………………… 4

Using cross multiplication method we get,

p = 1 and q = 1

p = 1/x

q = 1/y

x = 1 and y = 1

Question 43. 152x − 378y = − 74

− 378x + 152y = − 604

Sol. 152x − 378y = − 74 ……………………………. 1

− 378x + 152y = − 604 …………………………. 2

Adding the equations 1 and 2 , we obtain

− 226x − 226y = − 678

= x + y = 3 ……………………….. 3

Subtracting the equation 2 from equation 1, we obtain

530x + 530y = 530

x − y = 1 …………………………….4

Adding equations 3 and 4 we obtain,

2x = 4

= x = 2

Substituting the value of x in equation 3 we obtain y = 1

Question 44. 99x + 101y = 409

101x + 99y = 501

Sol. The given system of equation are :

99x + 101y = 409 ………………………….1

101x + 99y = 501…………………………….. 2

Adding equation 1 and 2 we get ,

99x + 101x + 101y + 99y = 49 + 501

= 200(x + y) = 1000

= x + y = 5 ……………………….. 3

Subtracting equation 1 from 2

101x − 99x + 99y − 101y = 501 − 499

= 2(x − y) = 2

= x − y = 1 ………………………………. 4

Adding equation 3 and 4 we get,

2x = 6

= x = 3

Putting x = 3 in equation 3 we get,

3 + y = 5

= y = 2

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

Question 45. 23x − 29y = 98

29x − 23y = 110

Sol. 23x − 29y = 98 ………………………………1

29x − 23y = 110 ………………………………… 2

Adding equation 1 and 2 we get,

= 6(x + y) = 12

= x + y = 2 ………………………3

Subtracting equation 1 from 2 we get,

52(x − y) = 208

= x − y = 4 ……………………………. 4

Adding equation 3 and 4 we get,

2x = 6

= x = 3

Putting the value of x in equation 4

3 + y = 2

= y = − 1

The Solution of the given system of equation is 3 and − 1 respectively.

Question 46. x − y + z = 4

x − 2y − 2z = 9

2x + y + 3z = 1

Sol. x − y + z = 4 …………………………..1

x − 2y − 2z = 9…………………………..2

2x + y + 3z = 1……………………………3

From equation 1

z = 4 − x + y

z = − x + y + 4

Subtracting the value of the z in equation 2 we get,

x − 2y − 2( − x + y + 4) = 9

= x − 2y + 2x − 2y − 8 = 8

= 3x − 4y = 17 …………………………….. 4

Subtracting the value of z in equation 3, we get,

2x + y + 3( − x + y + 4) = 1

= 2x + y + 3x + 3y + 12 = 1

= − x + 4y = − 11

Adding equation 4 and 5 we get,

3x − x − 4y + 4y = 17 − 11

= 2x = 6

= x = 3

Putting x = 3 in equation 4, we get,

9 − 4y = 17

= − 4y = 17 − 9

= y = − 2

Putting x = 3 and y = − 2 in z = − x + y + 4 , we get,

Z = − 3 − 2 + 4

= − 1

The Solution of the given system of equation are 3 , − 2 and − 1 respectively.

Question 47. x − y + z = 4

x + y + z = 2

2x + y − 3z = 0

Sol. x − y + z = 4 ………………………………….1

x + y + z = 2……………………………………..2

2x + y − 3z = 0………………………………………3

From equation 1

= z = − x + y + 4

Substituting z = − x + y + 4 in equation 2 , we get ,

= x + y + ( − x + y + 4) = 2

= x + y − x + y + 4 = 2

= 2y = 2

= y = 1

Substituting the value of z in equation 3

2x + y − 3( − x + y + 4) = 0

= 2x + y + 3x − 3y − 12 = 0

= 5x − 2y = 12 ………………………………. 4

Putting the y = − 1 in equation 4

5x − 2( − 1) = 12

5x = 10

= x = 2

Putting x = 2 and y = − 1 in z = − x + y + 4

Z = − 2 − 1 + 4

= 1

The Solution of the given system of equations are 2 , − 1 and 1 respectively.