Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x + 4y = 3 and 12y + 6x = 6

(ii) x = 2y and y = 2x

(iii) 3x + y - 3 = 0 and

The Condition for no solution is : (parallel lines)

(i) Yes.

Given pair of equations are,

2x+4y - 3 = 0 and 6x + 12y - 6 = 0

Comparing with ax+ by +c = 0;

Here, a₁ = 2, b₁ = 4, c₁ = - 3;

And a₂ = 6, b₂ = 12, c₂ = - 6;

a₁ /a₂ = 2/6 = 1/3

b₁ /b₂ = 4/12 = 1/3

c₁ /c₂ = - 3/ - 6 = 1/2

Here, a₁/a₂ = b₁/b₂ c₁/c₂, i.e parallel lines

Hence, the given pair of linear equations has no solution.

(ii) No.

Given pair of equations,

x = 2y and y = 2x

or x - 2y = 0 and 2x - y = 0;

Comparing with ax + by + c = 0;

Here, a₁ = 1, b₁ = - 2, c₁ = 0;

And a₂ = 2, b₂ = - 1, c₂ = 0;

a₁ /a₂ = 1/2

b₁ /b₂ = -2/-1 = 2

Here, a₁/a₂ b₁/b_2.

Hence, the given pair of linear equations has unique solution.

(iii) No.

Given pair of equations,

3x + y - 3 = 0

and

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = 1, c₁ = - 3;

And a₂ = 2, b₂ = 2/3, c₂ = - 2;

a₁ /a₂ = 2/6 = 3/2

b₁ /b₂ = 4/12 = 3/2

c₁ /c₂ = - 3/-2 = 3/2

Here, a₁/a₂ = b₁/b₂ = c₁/c₂, i.e coincident lines

Hence, the given pair of linear equations is coincident and having infinitely many solutions.