Solve the following pair of linear equations by the substitution method.
(i) \(x + y = 14,x – y = 4\)
(ii) \(s – t = 3,{{s}\over{3}} + {{t}\over{2}} = 6\)
(iii) \(3x – y = 3,9x - 3y = 9\)
(iv) \(0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3\)
(v) \(\sqrt{2}x + \sqrt{3}y = 0,\sqrt{3}x - \sqrt{8}y = 0\)
(vi) \({{3x}\over{2}} - {{5y}\over{3}} = -2,{{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}\)

Ans.(i)\(x + y = 14\)...........(i)
=> \(x – y = 4\)..............(ii)

Using equation (ii)

=> \(x = 4 + y\).............(iii)

On substituting (iii) in (i):

=>\(4 + 2y = 14\)
=>\(2y = 10\)
=>\(y = 5\)...........(iv)

On substituting (iv) in (i):
=>\(x + 5 = 14\)
=>\(x = 14 - 5\)
=>\(x = 9\)

So, the value of \(y = 5\) and \(x = 9\).

(ii)\(s – t = 3\)...........(i)
=> \({{s}\over{3}} + {{t}\over{2}} = 6\)..............(ii)

Using equation (i)
=> \(s = 3 + t\).............(iii)

On substituting (iii) in (ii):
=>\({{3 + t}\over{3}} + {{t}\over{2}} = 6\)
=>\({{6 + 2t + 3t}\over{6}} = 6\)
=>\(5t + 6 = 36\)
=>\(5t = 30\)
=>\(t = 6\)..............(iv)

On substituting (iv) in (i):
=>\(s – 6 = 3\)
=>\(s = 6 + 3\)
=>\(s = 9\)

So, the value of \(t = 6\) and \(s = 9\).

(iii) \(3x – y = 3,9x - 3y = 9\)
=> \(3x - y = 3\)..........(i)
=> \(9x - 3y = 9\)........(ii)

Using equation (i),
=> \(y = 3x - 3\)........(iii)

On substituting (iii) in (ii):
=> \(9x - 9x + 9 = 9\)
=> 9 = 9

This is always true.

Hence the given pair of equations have infinitely many solutions. The relation between them is given by :
\(y = 3x - 3\)

One such solution can be x = 1 and y = 0.

(iv)\(0.2x + 0.3y = 1.3\)...........(i)
=> \(0.4x + 0.5y = 2.3\)..............(ii)

Using equation (i)
=> \(0.2x = 1.3 - 0.3y\)
=>\(x = {{1.3 - 0.3y} \over {0.2}}\).............(iii)

On substituting (iii) in (ii):
=>\(0.4({{1.3 - 0.3y }\over {0.2}}) + 0.5y = 2.3\)
=>\(2.6 - 0.6y + 0.5y = 2.3\)
=>\(-0.1y = -0.3\)
=>\(y = 3\).............(iv)

On substituting (iv) in (i):

=>\(0.2x + 0.3 (3) = 1.3\)
=>\( 0.2x + 0.9 = 1.3\)
=>\(0.2x = 0.4 => x = 2\)

So, the value of \(y = 3\) and \(x = 2\).

(v)\(\sqrt{2}x + \sqrt{3}y = 0\)...........(i)
=> \(\sqrt{3}x - \sqrt{8}y = 0\)..............(ii)

Using equation (i)
=> \(x = y\sqrt{{-3}\over {2}}\).............(iii)

On substituting (iii) in (ii):
=>\(\sqrt{3}(y\sqrt{{-3}\over {2}}) - \sqrt{8}y =0\)
=>\({{-3y}\over {\sqrt 2}} - \sqrt{8}y =0\)
=>\(y({{-3}\over {\sqrt 2}} - \sqrt{8}) =0\)
=>\(y = 0\).............(iv)

On substituting (iv) in (iii):
=>\(x = 0\)

So, the value of \(y = 0\) and \(x = 0\).

(vi)\({{3x}\over{2}} - {{5y}\over{3}} = -2\)...........(i)
=> \({{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}\)..............(ii)

Using equation (ii)
=> \(x = ({{13}\over{6}} - {{y}\over{2}}) × 3\)
=> \(x = {{13}\over{2}} - {{3y}\over{2}}\).............(iii)

On substituting (iii) in (i):
=>\({3({{13}\over{2}} - {{3y}\over{2}})\over{2}} - {{5y}\over{3}} = -2\)
=>\({{39}\over{4}} - {{9y}\over{4}} - {{5y}\over{3}} = -2 \)
=>\({{-27y -20y}\over{12}} = -2 - {{39}\over{4}}\)
=>\({{-47y}\over{12}} = {{-8-39}\over{4}} \)
=>\({{-47y}\over{12}} = {{-47}\over{4}} \)
=>\(y = 3\)

On substituting (iv) in (ii):
=>\({{x}\over{3}} + {{3}\over{2}} = {{13}\over{6}}\)
=>\({{x}\over{3}} = {{13}\over{6}} - {{3}\over{2}}\)
=>\({{x}\over{3}} = {{2}\over{3}}\)
=>\(x = 2\)

So, the value of \(y = 3\) and \(x = 2\).