Solve the following pair of linear equations by the elimination method and the substitution method:
(i) \(x + y = 5,2x – 3y = 4\)
(ii) \(3x + 4y = 10, 2x – 2y = 2\)
(iii) \(3x - 5y – 4 = 0, 9x = 2y + 7\)
(iv) \({{x}\over{2}} + {{2y}\over{3}} = -1, x - {{y}\over{3}} = 3\)

Ans.(i) \(x + y = 5\) … (1)
\(2x – 3y = 4\) … (2)

Elimination method:

On multiplying equation (1) by 2, we get equation (3)

\(2x + 2y = 10\) … (3)
\(2x - 3y = 4\) … (2)

On subtracting equation (2) from (3), we get
\(5y = 6 \)
=> \( y = {{6}\over{5}}\)

On putting value of y in (1), we get
\(x + {{6}\over{5}} = 5\)
\( x = 5 - {{6}\over{5}} = {{19}\over{5}}\)

Therefore, x = \({{6}\over{5}}\) and y = \({{19}\over{5}}\)

Substitution method:
\(x + y = 5\) … (1)
\(2x - 3y = 4\) … (2)

From equation (1), we get,
\(x = 5 - y\)

On putting this in equation (2), we get
\(2 (5 - y) - 3y = 4\)
=>\( 10 - 2y - 3y = 4\)
=>\( 5y = 6\)
=>\( y = {{6}\over{5}}\)

On putting value of y in (1), we get
\(x = 5 - {{6}\over{5}} = {{19}\over{5}}\)

Therefore, y = \({{6}\over{5}}\) and x = \({{19}\over{5}}\)

(ii) \(3x + 4y = 10\) … (1)
\(2x – 2y = 2\) … (2)

Elimination method:

On multiplying equation (2) by 2, we get equation (3)

\(4x - 4y = 4\) … (3)
\(3x + 4y = 10\) … (1)

On adding equation (1) and (3), we get
\(7x = 14 => x = 2\)

On putting value of x in (1), we get
\(3 (2) + 4y = 10\)
\( 4y = 10 – 6 = 4\)
\( => y = 1\)

Therefore, x = \(2\) and y = \(1\)

Substitution method:
\(3x + 4y = 10\) … (1)
\(2x - 2y = 2\) … (2)

From equation (2), we get,
\(2x = 2 + 2y\)
=> \(x = 1 + y \)..........(3)

On putting this in equation (1), we get
\(3 (1 + y) + 4y = 10\)
=>\( 3 + 3y + 4y = 10\)
=>\( 7y = 7 => y = 1\)

On putting value of y in (3), we get
\( x = 1 + 1 = 2\)

Therefore, x = \(2\) and y = \(1\)


(iii) \(3x - 5y – 4 = 0\) … (1)
\(9x = 2y + 7\)… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

\(9x - 15y – 12 = 0\)… (3)
\(9x - 2y – 7 = 0\)… (2)

On subtracting (2) from (3), we get
\(-13y – 5 = 0\)
=>\( -13y = 5\)
=> \(y = {{-5} \over {13}}\)

On putting value of y in (1), we get
\(3x – 5({{-5} \over {13}}) - 4 = 0\)
=>\( 3x = 4 - {{25} \over {13}}= {{52 - 25} \over {13}} = {{27} \over {13}}\)
=> \(x = {{27} \over {13(3)}} = {{9} \over {13}}\)
Therefore, x = \({{9} \over {13}}\) and y = \({{-5} \over {13}}\)

Substitution Method:
\(3x - 5y – 4 = 0\) … (1)
\(9x = 2y + 7\)… (2)

From equation (1), we can say that
\(3x = 4 + 5y => x = {{4 + 5y} \over {3}}\)

On putting this in equation (2), we get
\(9({{4 + 5y} \over {3}}) - 2y = 7\)
=>\( 12 + 15y - 2y = 7\)
=>\( 13y = -5 \)
\(y = {{-5} \over {13}}\)

On putting value of y in (1), we get
\(3x – 5 ({{-5} \over {13}})= 4\)
=>\( 3x = 4 - {{25} \over {13}} = {{52 - 25} \over {13}} = {{27} \over {13}}\)
=>\( x = {{27} \over {13(3)}} = {{9} \over {13}}\)

Therefore, x = \({{9} \over {13}}\) and y = \({{-5} \over {13}}\)

(iv)\({{x}\over{2}} + {{2y}\over{3}} = -1\) … (1)
\( x - {{y}\over{3}} = 3\)… (2)

Elimination method:

On multiplying equation (2) by 2, we get (3)

\( 2x - {{2y}\over{3}} = 6\)… (3)
\({{x}\over{2}} + {{2y}\over{3}} = -1\)… (1)

Adding (3) and (1), we get

\({{5x}\over{2}} = 5 => x = 2\)

On putting value of x in (2), we get
\(2 - {{y}\over{3}} = 3\)
=>\( y = -3\)

Therefore, x = 2 and y = -3

Substitution method:

\({{x}\over{2}} + {{2y}\over{3}} = -1\) … (1)
\( x - {{y}\over{3}} = 3\)… (2)

From equation (2), we can say that
=> \(x = 3 + {{y}\over{3}} = {{9 + y}\over{3}}\)

Putting this in equation (1), we get
=>\({{9 + y}\over{6}} + {{2y}\over{3}} = -1\)
=>\({{9 + y + 4y}\over{6}} = -1 \)
=>\(5y + 9 = -6\)
=>\(5y = -15 \)
=>\(y = -3\)

Putting value of y in (1), we get
=>\({{x}\over{2}} + {{2(-3)}\over{3}} = -1\)
=>\(x = 2\)

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