Question

Solve the following pair of linear equation by the elimination method and the substitution method.

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

Answer 1

`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)

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.