Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination method:

Multiplying equation (1) by 2, we get equation (3)

2x + 2y = 10 … (3)

2x − 3y = 4 … (2)

Subtracting equation (2) from (3), we get

5y = 6⇒ y = 6/5

Putting value of y in (1), we get

x + 6/5= 5

⇒ x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

Substitution method:

x + y = 5 … (1)

2x − 3y = 4 … (2)

From equation (1), we get,

x = 5 − y

Putting this in equation (2), we get

2 (5 − y) − 3y = 4

⇒ 10 − 2y − 3y = 4

⇒ 5y = 6 ⇒ y = 6/5

Putting value of y in (1), we get

x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

(ii) 3x + 4y = 10… (1)

2x – 2y = 2… (2)

Elimination method:

Multiplying equation (2) by 2, we get (3)

4x − 4y = 4 … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14⇒ x = 2

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)

Putting this in equation (1), we get

3 (1 + y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 7⇒ y = 1

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)

Subtracting (2) from (3), we get

−13y – 5 = 0

⇒ −13y = 5

⇒ y = −5/13

Putting value of y in (1), we get

3x – 5 (−5/13)− 4 = 0

⇒ 3x = 4 −

⇒ x =

Therefore, x = 9/13and y = -5/13

Substitution Method:

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

From equation (1), we can say that

3x = 4 + 5y⇒ x =

Putting this in equation (2), we get

9 − 2y = 7

⇒ 12 + 15y − 2y = 7

⇒ 13y = −5 ⇒ y = -5/13

Putting value of y in (1), we get

3x – 5 (-5/13)= 4

⇒ 3x = 4 −

⇒ x =

Therefore, x = 9/13 and y = -5/13

(iv) x/2 + 2y/3 = -1 ....(1)

x - y/3 = 3… (2)

Elimination method:

Multiplying equation (1) by 2, we get

… (3)

substracting equation (2)from (3) we get

Putting value of x in (2), we get

x +  = -2

=x-4=-2
=x=2

Substitution method:

From equation (2), we can say that

Putting this in equation (1), we get