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

`x/2 + (2y)/3 = -1 and x - y /3 = 3`

#### Solution

x/2 + 2y/3 = - 1 and x – y/3 = 3

By elimination method

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

x – y/3 = 3 ... (ii)

Multiplying equation (i) by 2, we get

x + 4y/3 = - 2 ... (iii)

x – y/3 = 3 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y/3 = -5

Dividing by 5 and multiplying by 3, we get

y = -15/5

y = - 3

Putting this value in equation (ii), we get

x – y/3 = 3 ... (ii)

x – (-3)/3 = 3

x + 1 = 3

x = 2

Hence our answer is x = 2 and y = −3.

By substitution method

x – y/3 = 3 ... (ii)

Add y/3 both side, we get

x = 3 + y/3 ... (iv)

Putting this value in equation (i) we get

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

(3+ y/3)/2 + 2y/3 = -1

3/2 + y/6 + 2y/3 = - 1

Multiplying by 6, we get

9 + y + 4y = - 6

5y = -15

y = - 3

Hence our answer is x = 2 and y = −3.