Solve the following pair of linear equations by the substitution method.

(i)	(ii)
(iii)	(iv)
(v)	(vi)

i) x + y = 14...............(i)

x - y = 4....................(ii)

From equation (i) we get,

x = 14 - y .........(iii)

Putting value of x in equation (ii) we get,

(14 - y) - y = 4

= 14 - 2y = 4

= 2y = 10

= y =

Putting value of y in equation (iii) we get,

x = 14 -5 = 9

Hence, x = 9 and y = 5

ii) s - t = 3......................(i)

=

From equation (i) we get,

s = t + 3...............(iii)

Putting value of x from (iii) to (ii)

=

= 2t+6+3t = 36

= 5t = 30

= t =

Putting value of t in equation (iii) , we get,

s = 6 + 3 = 9

Hence, s = 9, t = 6

iii) 3x - y = 3..................(i)

9x - 3y = 9......................(ii)

From equation (i) , we get,

y = 3x - 3...............(iii)

Putting value of y from equation (iii) to equation (ii)

9x - 3(3x - 3) = 9

= 9x - 9x + 9 = 9

= 9 = 9

This is always true, and pair of these equations have infinite possible solutions

Therefore one possible solution is = x = 1 and y = 0

iv) 0.2x + 0.3y = 1.3 .....................(i)

0.4 x + 0.5 y = 2.3 ........................(ii)

From equation (i) . we get,

x =

Putting value of x in equation (ii) we get,

(6.5 - 1.5y) × 0.4x + 0.5y = 2.3

2.6 - 0.6y + 0.5y = 2.3

y =

Putting value of y in equation (iii) we get,

x = 6.5 - 1.5 × 3 = 6.5 - 4.5 = 2

Hence, x = 2 and y = 3

v)

=

From equation (i) , we get,

x =

Putting value of x in equation (ii). we get,

=

=

=

so, y = 0

Putting value of y in equation (iii) we get.

x = 0

Hence, x = 0 and y = 0

vi)

=

From equation (i) we get,

9x - 10y = -12

x =

Putting this value of x in equation (ii), we get,

=

=

= 47y = 117 + 24

= 47y = 141

y =

Putting value of y in (iii)

x =

Hence, x = 2 and y = 3