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

(i) x + y = 14 …(1)

x – y = 4 … (2)

x = 4 + y from equation (2)

Putting this in equation (1), we get

4 + y + y = 14

⇒ 2y = 10⇒ y = 5

Putting value of y in equation (1), we get

x + 5 = 14

⇒ x = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s - t = 3......................(i)
= s/3 + t/2 = 6 ................(ii)
From equation (i) we get,
s = t + 3...............(iii)
Putting value of  x from (iii) to (ii)
=
= 2t+6+3t = 36
= 5t = 30
= t = 6
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,

0.2x = 1.3 − 0.3y
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

−0.1y = −0.3 ⇒ y = 3
y = 3
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)  √2x+ √3y = 0 .....(i)

√3x – √8y = 0.........(ii)
From equation (i) , we get,
x =
Putting value of x  in equation (ii). we get,

Putting value of  y  in equation (iii) we get.
x = 0
Hence, x = 0   and  y = 0
(vi)  3x/2 - 5y/3 = -2
x/3 + y/2 = 13/6
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 = 3
Putting value of y in (iii)
 
Hence, x = 2  and  y = 3