**Exercise 3.3**

**1.Solve the following pair of linear equations by the substitution method **:

**Sol.** (i) The given system of equations is

x +3 = 14 …(1)

#### and, x-y = 4 …(2)

#### Substituting y = 14-x in (2), we get

x – (14 -x) = 14 => x-14 x = 4

=> 2x = 4 + 14 => 2x = 18

=> x= 9

Putting x = 9 in (1), we get

9+y = 14 => y = 14 – 9 => y =5

Hence, the solution of the given system of equations is

#### x= 9, y = 5

#### (ii) The given system of equations is

#### s-t = 3 …(1)

#### From (1), s =3+1

Substituting 8 = 3 + t in (2), we get

#### => 2(3+*t*) +3*t* =36

#### => 6 + 2*t* + 3*t* = 36 => 5*t* = 30 =›* t *= 6

Putting *t* = 6 in (1), we get

s – 6 = 3 => s = 3 + 6 = 9

Hence, the solution of the given system of equations is

s= 9,*t*= 6

*(iii)* The given system of equations is

3x – y = 3 …(1)

and, 9x – 3y = 9 … (2)

From (1), y = 3x – 3

Substituting y = 3x – 3 in (2), we get

9x – 3(3x – 3) = 9 => 9x – 9x + 9 = 9

=> 9 = 9

This statement is true for all values of *x*. However, we do not get a specific value of *x* as a solution. So, we cannot obtain a specific value of *y*. This situation has arisen because both the given equations are the same.

* Equations (1) and (2) have infinitely many solutions.

*(iv)* The given system of equations is

0.2x + 0.3y = 1.3 => 2x + 3y = 13 …(1)

and, 0.4x + 0.5y = 2.3 => 4x + 5y = 23 …(2)

**(vi) The given system of equations is**

**2.Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence**

find the value of ‘*m*‘ for which y = mx + 3.

find the value of ‘

*m*‘ for which y = mx + 3.