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RD Sharma Solutions for Class 10 Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Download PDF

Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.4

Question: 1

Solve the system of equations:

x + 2y + 1 = 0 and 2x – 3y – 12 = 0

Solution:

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 = 3

And,

= y = – 2

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

Question: 2

Solve the system of equations:

3x + 2y + 25 = 0, 2x + y + 10 = 0

Solution:

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

And,

y = – 20

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

Question: 3

Solve the system of equations:

2x + y = 35, 3x + 4y = 65

Solution:

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 = 15

And,

y = 5

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

Question: 4

Solve the system of equations:

2x – y – 6 = 0, x – y – 2 = 0

Solution:

2x – y = 6 …. (i)

x – y = 2 ….. (ii)

Here a₁ = 2, b₁ = -1, c₁ = 6

a ₂= 1, b₂ = -1, c₂ = 2

By cross multiplication method,

Now,

x = 4

And,

 y = 2

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

Question: 5

Solve the system of equations:

Solution:

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

X = – 1/2

y = –1/4

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

Question: 6

Solve the system of equations:

ax + by = a – b, bx – ay = a + b

Solution:

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.

Question: 7

Solve the system of equations:

x + ay – b = 0, ax – by – c = 0

Solution:

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 isrespectively.

Question: 8

Solve the system of equations:

ax + by = a²

bx + ay = b²

Solution:

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.

Question: 9

Solve the system of equations:

Solution:

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,

x + y = 5 …. (i)

And,

v = 1

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.

Question: 10

Solve the system of equations:

Solution:

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 = 2

And,

v = 3

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

Question: 11

Solve the system of equations:

Solution:

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,

x + y = 19 ….. (i)

And,

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.

Question: 12

Solve the system of equations:

Solution:

a₂ = a, b₂ = – b, c₂ = b² – a²

By cross multiplication method

x = a

y = b

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

Question: 13

Solution:

By cross multiplication method

x = a²

y = b²

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

Question: 14

ax + by = a² + b²

Solution:

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

By cross multiplication method

x = a

y = b

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

Question: 15

2ax + 3by = a + 2b

3ax + 2by = 2a + b

Solution:

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,

The solutions of the system of equations are respectively.

Question: 16

5ax + 6by = 28

3ax + 4by = 18

Solution:

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,

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

Question: 17

(a + 2b)x + (2a – b)y = 2

(a – 2b)x + (2a + b)y = 3

Solution:

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

Question: 18

Solve the system of equations:

Solution:

The given systems of equations are:

From equation (i)

From equation (ii)

x + y – 2a² = 0

a₂ = 1, b₂ = 1, c₂ = – 2a²

By cross multiplication method:

The solutions of the given system of equations arerespectively.

Question: 19

Solve the system of equations:

Solution:

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.

Question: 20

Solve the system of equations:

(a – b)x +(a + b)y = 2a² – 2b²(a + b)(x + y) = 4ab

Solution:

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 arerespectively.

Question: 21

Solve the system of equations:

a²x + b²y = c²

b²x + a²y = d²

Solution:

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,

The solution of the given system of equations arerespectively.

Question: 22

Solve the system of equations:

2(ax – by + a + 4b = 0

2(bx + ay) + b – 4a = 0

Solution:

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

y = 2

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

Question: 23

Solve the system of equations:

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

6(bx – ay) = 3b – 2a

Solution:

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

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

Question: 24

Solve the system of equations:

Solution:

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

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

Question: 25

Solve the system of equations:

mx – my = m² + n²

x + y = 2m

Solution:

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

y = m – n

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

Question: 26

Solve the system of equations:

ax – by = 2ab

Solution:

The given pair of equations are:

ax – by = 2ab ….. (ii)

Here, a₂ = a, b₂ = – b, c₂ = – (2ab)

By cross multiplication method

x = b

y = – a

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

Question: 27

Solve the system of equations:

x + y – 2ab = 0 ……. (ii)

Solution:

x + y – 2ab = 0 ……. (ii)

By cross multiplication method

x = ab

y = ab

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

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