Ex-3.4 Pair Of Linear Equations In Two Variables, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q.1: x + 2y + 1 = 0 and 2x – 3y – 12 = 0

Sol.

x + 2y + 1 = 0 …………………………………….(i)

2x-3y-12 = 0………………………………….. (ii)

Here a₁ = 1 , b₁ = 2 , c₁ = 1

a₂ = 2 , b₂ = -3 , c₂ = -12

By cross multiplication method,

Now,

x /− 21 = 1 /− 7

= x = 3

And,

− y /− 14 = 1 /− 7

= y = -2

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

Q.2: 3x + 2y + 25 = 0, 2x + y + 10 = 0

Sol. 3x + 2y + 25 = 0 …………………………………….(i)

2x + y + 10 = 0………………………………….. (ii)

Here a₁ = 3 , b₁ = 2 , c₁ = 25

a₂ = 2 , b₂ = 1 , c₂ = 10

By cross multiplication method,

Now,

x/ − 5 = 1 /− 1

= x = 5

And,

− y /− 20 = 1/ − 1

= y = -20

The solution of the given system of equation is 5 and -20 respectively.

Q.3: 2x + y = 35, 3x + 4y = 65

Sol. 2x + y = 35 ……………………………….(i)

3x + 4y = 65…………………………….. (ii)

Here a₁ = 2 , b₁ = 1 , c₁ = 35

a₂ = 3 , b₂ = 4 , c₂ = 65

By cross multiplication method,

Now,

x/75 = 1/5

= x = 15

And,

− y /− 25 = 1/5

= y = 5

The solution of the given system of equation is 15 and 5 respectively.

Q.4: 2x – y – 6 = 0, x – y – 2 = 0

Sol. 2x-y = 6 ……………………………….(i)

x-y = 2…………………………….. (ii)

Here a₁ = 2 , b₁ = -1 , c₁ = 6

a₂ = 1 , b₂ = -1 , c₂ = 2

By cross multiplication method,

Now,

x/ − 4 = 1/ − 1

= x = 4

And,

− y/2 = 1 /− 1

= y = 2

The solution of the given system of equation is 4 and 2 respectively.

Q5: 

Sol.  

=  1/x + 1/y = 2 ……………………………….. (i)

=  1/x − 1/y = 6 ………………………………… (ii)

Taking 1/x  = u

Taking 1/y  = v

= u + v = 2 …………………………… (iii)

= u-v = 6 ……………………………. (iv)

By cross multiplication method,

Now,

u/4 = 1/ − 2

= u = -2

And,

− v/ − 8 = 1/ − 2

= v = 4

1/u  = x = − 1/2

1/v  = y = 1/4

The solution of the given system of equation is  − 1/2  and 1/4 respectively.

Q.6: ax + by = a-b, bx-ay = a + b

Sol. ax + by = a-b……………………………….(i)

bx-ay = a + b…………………………….. (ii)

Here a₁ = a , b₁ = b , c₁ = a-b

a₂ = b , b₂ = -a , c₂ = a + b

By cross multiplication method,

Now,

= x = 1

And,

= y = -1

The solution of the given system of equation is 1 and -1 respectively.

Q.7: x + ay-b = 0, ax-by-c = 0

Sol. x + ay-b = 0 ……………………………….. (i)

ax-by-c = 0………………………………. (ii)

Here a₁ = 1 , b₁ = a , c₁ = -b

a₂ = a , b₂ = -b , c₂ = -c

By cross multiplication method,

Now,

And,

The solution of the given system of equation is  respectively.

Q8 ax + by = a²

bx + ay = b²

Sol. ax + by = a²……………………………….(i)

bx + ay = b²…………………………….. (ii)

Here a₁ = a , b₁ = b , c₁ = a²

a₂ = b , b₂ = a , c₂ = b²

By cross multiplication method,

Now,

And,

The solution of the given system of equation is   respectively.

Q9

Sol.

The given system of equations are :

5u-2v = -1

15u + 7v = 10

Here a₁ = 5, b₁ = -2 , c₁ = 1

a₂ = 15 , b₂ = 7 , c₂ = -10

By cross multiplication method,

Now,

u/13 = 1 /− 65

= u =  1/5

1/u =   = x + y

= x + y = 5 ………………………..(i)

And,

− v / − 65 = 1/ − 65

= v = 1

1v =   = x-y

= x-y = 1 ………………………… (ii)

Adding equation (i) and (ii)

2x = 6

= x = 3

Putting the value of x in equation (i)

3 + y = 5

= y = 2

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

Q10

2/x + 3/y = 13

5/x − 4/y = − 2

Sol.

Let 1/x  = u

Let 1/y  = v

The given system of equations becomes:

2u + 3v = 13 ……………………………… (i)

5u-4v = -2…………………………………. (ii)

By cross multiplication method,

Now,

u /− 46 = 1/ − 23

= u = 2

1/u =   = 1/x

= x = 12

And,

− v/69 = 1 /− 23

= v = 3

1/v =  1/y

= y = 13

The solutions of the given system of equations are 1/2 and 1/3 respectively.

Q11

Sol.

The given system of equations are :

57u + 6v = 5

38u + 21v = 9

Here a₁ = 57, b₁ = 6 , c₁ = -5

a₂ = 38 , b₂ = 21 , c₂ = -9

By cross multiplication method,

Now,

u/51 = 1/969

= u =  1/19

1/u  = x + y

= x + y = 19 ………………………..(i)

And,

− v/ − 323 = 1/969

= v = 1/3

1/v =   = x-y

= x-y = 3 ………………………… (ii)

Adding equation (i) and (ii)

2x = 22

= x = 11

Putting the value of x in equation (i)

11 + y = 19

= y = 8

The solution of the given system of equation is 11 and 8 respectively.

Q12

x/a − y/b = 2

ax-by = a²-b²

Sol.

a₁ = 1/a , Let b₁ = 1/b, Let c₁ = -2

a₂ = a ,  b₂ = -b,  c₂ = b²-a²

By cross multiplication method

Now, 

x = a

and, 

= y = b

Hence the solution of the given system of equation are a and b respectively.

Q13

Sol.

Here, a₁ = 1/a , Let b₁ = 1/b, Let c₁ = -(a + b)

a₂ = 1/a2,  b₂ = 1/b2,  c₂ = -2

By cross multiplication method

Now, 

= x = a²

= y = b²

The solution of the given system of equation are  a²  and b² respectively.

Q14

x/a = y/b

ax + by = a² + b²

Sol. Here, a₁ = 1a , Let b₁ = 1b,  c₁ = 0

Here, a₁ = a ,  b₂ = b, Let c₁ = -(a² + b²)

By cross multiplication method

Now, 

= x = a

And 

= y = b

The solution of the given system of equations are  a and b respectively.

Q15

2ax + 3by = a + 2b

3ax + 2by = 2a + b

Sol. The given system of equation is

2ax + 3by = a + 2b …………………………… (i)

3ax + 2by = 2a + b…………………………….. (ii)

Here a₁ = 2a, b₁ = 3b , c₁ = -(a + 2b)

a₂ = 3a , b₂ = 2b , c₂ = -(2a + b)

By cross multiplication method

Now,

And, 

The solutions of the system of equations are 

Q16

5ax + 6by = 28

3ax + 4by = 18

Sol.

The systems of equations are:

5ax + 6by = 28 …………………………. (i)

3ax + 4by = 18……………………………. (ii)

Here a₁ = 5a, b₁ = 6b , c₁ = -(28)

a₂ = 3a , b₂ = 4b , c₂ = -(18)

By cross multiplication method

Now,

x/4b = 1/2ab

= x =  2a

And,  − y/ − 6a = 1/2ab

= y = 3/b

The solution of the given system of equation is 2/a and 3/b.

Q17

(a + 2b)x + (2a-b)y = 2

(a-2b)x + (2a + b)y = 3

Soln.

The given system of equations are :

(a + 2b)x + (2a-b)y = 2 ………………………. (i)

(a-2b)x + (2a + b)y = 3………………………….. (ii)

Here a₁ = a + 2b, b₁ = 2a-b , c₁ = -(2)

a₂ = a-2b , b₂ = 2a + b , c₂ = -(3)

By cross multiplication method:

The solution of the system of equations are = x =  

And = y =  respectively.

Q18

x + y = 2a²

Sol. The given systems of equations are:

x + y = 2a²

From equation (i)

From equation (ii)

X + y-2a² = 0

Here  

a₂ = 1, b₂ = 1 , c₂ = -2a²

By cross multiplication method:

The solutions of the given system of equations are  respectively.

Q19

bx + cy = a + b

Sol. The system of equation is given by :

bx + cy = a + b ……………………………………. (i)

From equation (i)

bx + cy-(a + b) = 0

From equation (ii)

= 2abx-2acy-2a(a-b) = 0 …………………………. (iv)

By cross multiplication

= x = a/b

And,

= y = b/c

The solution of the system of equations are a/b and b/c

Q20

(a-b)x + (a + b)y = 2a²-2b²

(a + b)(x + y) = 4ab

Soln.

The given system of equations are :

(a-b)x + (a + b)y = 2a²-2b² ………………………….. (i)

(a + b)(x + y) = 4ab …………………………. (ii)

From equation (i)

(a-b)x + (a + b)y-2a²-2b²  = 0

= (a-b)x + (a + b)y-2(a²-b²) = 0

From equation (ii)

(a-b)x + (a-b)y-4ab = 0

Here, a₁ = a-b , b₁  = a + b , c₁ = -2(a² + b²)

Here, a₂ = a + b , b₂  = a + b , c₂ = -4ab

By cross multiplication method

Now,

The solution of the system of equations are respectively.

Q21

a²x + b²y = c²

b²x + a²y = d²

Sol.

The given system of equations are :

a²x + b²y = c² ………………………………….. (i)

b²x + a²y = d²……………………………………… (ii)

Here, a₁ = a² , b₁  = b² , c₁ = -c²

Here, a₂ = b² , b₂  = a² , c₂ = -d²

By cross multiplication method

Now,

And  = 

The solution of the given system of equations are  respectively.

Q23

2(ax-by + a + 4b = 0

2(bx + ay) + b-4a = 0

Sol. The given system of equation may be written as :

2(ax-by + a + 4b = 0 …………………….. (i)

2(bx + ay) + b-4a = 0……………………. (ii)

Here, a₁ = 2a , b₁  = -2b , c₁ = a + 4b

Here, a₂ = 2b , b₂  = 2a , c₂ = b-4a

By cross multiplication method

Now, 

= x = -1/2

 x − 2b2 − 2a2 = 14a2 + 4b2

= x =   − 12

And,  − y − 8a2 − 8b2 = 14a2 + 4b2

= y = 2

The solution of the given pair of equations are  − 1/2 and 2 respectively.

Q24

6(ax + by) = 3a + 2b

6(bx-ay) = 3b-2a

Sol. The systems of equations are

6(ax + by) = 3a + 2b ………………………… (i)

6(bx-ay) = 3b-2a …………………………. (ii)

From equation (i)

6ax + 6by-(3a + 2b) = 0  ……………………… (iii)

From equation (ii)

6bx-6ay-(3b-2a) = 0 …………………………… (iv)

Here, a₁ = 6a , b₁  = 6b , c₁ = -(3a + 2b)

Here, a₂ = 6b , b₂  = -6a , c₂ = -(3b-2a)

By cross multiplication method

Now, 

= x =  1/2

And , 

= y = 1/3

The solution of the given pair of equations are 1/2 and 1/3 respectively.

Q25

Sol. The given systems of equations are

Taking 1/x  = u

Taking 1/y = v

The pair of equations becomes:

a²u-b²v = 0

a²bu + b²av-(a + b) = 0

Here, a₁ = a² , b₁  = -b² , c₁ = 0

Here, a₂ = a²b , b₂  = b²a , c₂ = -(a + b)

By cross multiplication method

Now, 

= x =  1/a²

And, 

= y =  1/b²

The solution of the given pair of equations are 1/a² and  1/b² respectively.

Q26

mx-my = m² + n²

x + y = 2m

Sol. mx-my = m² + n²………………………………… (i)

x + y = 2m…………………………………………….. (ii)

Here, a₁ = m , b₁  = -n , c₁ = -(m² + n²)

Here, a₂ = 1 , b₂  = 1 , c₂ = -(2m)

By cross multiplication method

= x = m + n

And, 

= y = m-n

The solutions of the given pair of equations are m + n and m-n respectively.

Q27

ax/b − by/a = a + b

ax-by = 2ab

Sol. The given pair of equations are:

ax/b − by/a = a + b ……………………….. (i)

ax-by = 2ab …………………………….. (ii)

Here, a₁ =  a/b , b₁  = – b/a , c₁ = -(a + b)

Here, a₂ = a , b₂  = – b, c₂ = -(2ab)

By cross multiplication method

= x = b

And , 

= y = -a

The solution of the given pair of equations are b and –a respectively.

Q28

X + y-2ab = 0

Sol.

………………….. (i)

X + y-2ab = 0…………………………………………. (ii)

Here, a₁ =  b/a , b₁  =  a/b , c₁ = -(a² + b²)

Here, a₂ = 1 , b₂  = -1, c₂ = -(2ab)

By cross multiplication method

Now, 

= x = ab

And, 

= y = ab

The solutions of the given pair of equations are ab and ab respectively.