Conditions for pair of linear equations to be consistent are:

a₁/a₂ ≠ b₁/b_2. [unique solution]

a₁/a₂ = b₁/b₂ = c₁/c₂ [coincident or infinitely many solutions]

(i) No.

The given pair of linear equations

– 3x – 4y – 12 = 0 and 4y + 3x – 12 = 0

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = – 3, b₁ = – 4, c₁ = – 12;

a₂ = 3, b₂ = 4, c₂ = – 12;

a₁ /a₂ = – 3/3 = – 1

b₁ /b₂ = – 4/4 = – 1

c₁ /c₂ = – 12/ – 12 = 1

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) Yes.

The given pair of linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 3/5, b₁ = – 1, c₁ = – ½;

a₂ = 1/5, b₂ = 3, c₂ = – 1/6;

a₁ /a₂ = 3

b₁ /b₂ = – 1/ – 3 = 1/3

c₁ /c₂ = 3

Here, a₁/a₂ ≠ b₁/b_2.

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

(iii) Yes.

The given pair of linear equations –

2ax + by –a = 0 and 4ax + 2by – 2a = 0

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 2a, b₁ = b, c₁ = – a;

a₂ = 4a, b₂ = 2b, c₂ = – 2a;

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = ½

Here, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent

(iv) No.

The given pair of linear equations

x + 3y = 11 and 2x + 6y = 11

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 1, b₁ = 3, c₁ = 11

a₂ = 2, b₂ = 6, c₂ = 11

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = 1

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

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