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# 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 $$x = 2$$.