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

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# 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 a= 1, b= 2, c= 1

a= 2, b= -3, c= – 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 a= 3, b= 2, c= 25

a= 2, b= 1, c= 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 a= 2, b= 1, c= 35

a= 3, b= 4, c= 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 a= 2, b= -1, c= 6

2= 1, b= -1, c= 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 a= a, b= b, c= a – b

a= b, b= -a, c= 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 a= 1, b= a, c= – b

a= a, b= – b , c= – c

By cross multiplication method,

Now,

And,

The solution of the given system of equation is respectively.

### Question: 8

Solve the system of equations:

ax + by = a2

bx + ay = b2

### Solution:

ax + by = a…. (i)

bx + ay = b….. (ii)

Here a= a, b= b, c= a2

a= b, b= a, c= 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 a= 5, b= – 2, c= 1

a= 15, b= 7, c= 10

By cross multiplication method,

Now,

x + y = 5 …. (i)

And,

v = 1

x – y =  1 …… (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 a= 57, b= 6, c= – 5

a= 38, b= 21, c= – 9

By cross multiplication method,

Now,

x + y = 19 ….. (i)

And,

x – y = 3 … (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:

a= a, b= – b, c= b– a2

By cross multiplication method

x = a

y = b

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

### Solution:

By cross multiplication method

x = a2

y = b2

The solution of the given system of equation are a2  and brespectively.

ax + by = a+ b2

### Solution:

Here, a= a, b= b, Let c= – (a+ 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 a= 2a, b= 3b, c= – (a + 2b)

a= 3a, b= 2b, c= – (2a + b)

By cross multiplication method

Now,

The solutions of the system of equations are respectively.

5ax + 6by = 28

3ax + 4by = 18

### Solution:

The systems of equations are:

5ax + 6by = 28 …. (i)

3ax + 4by = 18  …. (ii)

Here a= 5a, b= 6b, c= – (28)

a= 3a, b= 4b, c= – (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 a= a + 2b, b= 2a – b, c= – (2)

a= a – 2b, b= 2a + b, c= – (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 – 2a= 0

a= 1, b= 1, c= – 2a2

By cross multiplication method:

The solutions of the given system of equations are respectively.

### 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 = 2a– 2b2(a + b)(x + y) = 4ab

### Solution:

The given system of equations are:

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

(a + b)(x + y) = 4ab      …… (ii)

From equation (i)

(a – b)x + (a + b)y – 2a– 2b= 0

= (a – b)x + (a + b)y – 2(a– b2) = 0

From equation (ii)

(a – b)x + (a – b)y – 4ab = 0

Here, a= a – b, b= a + b, c= – 2(a+ b2)

Here, a= a + b, b= a + b, c= – 4ab

By cross multiplication method

Now,

The solution of the system of equations are respectively.

### Question: 21

Solve the system of equations:

a2x + b2y = c2

b2x + a2y = d2

### Solution:

The given system of equations are:

a2x + b2y = c….. (i)

b2x + a2y = d…… (ii)

Here, a= a2, b= b2, c= – c2

Here, a= b2, b= a2, c= – d2

By cross multiplication method

Now,

The solution of the given system of equations are respectively.

### 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, a= 2a, b= -2b, c= a + 4b

Here, a= 2b, b= 2a, c= 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, a= 6a, b= 6b, c= – (3a + 2b)

Here, a= 6b, b= -6a, c= – (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, a= a2, b= -b2, c= 0

Here, a= a2b, b= b2a, c= – (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 = m+ n2

x + y = 2m

### Solution:

mx – my = m+ n…… (i)

x + y = 2m …….. (ii)

Here, a= m, b= -n, c= – (m+ n2)

Here, a= 1, b= 1, c= – (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, a= a, b= – b, c= – (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.

All Chapter RD Sharma Solutions For Class10 Maths

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