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Consider the following set of equations

$$x+2y=5\\ 4x+8y=12\\ 3x+6y+3z=15$$

This set

1. has unique solution

2. has no solution

3. has finite number of solutions

4. has infinite number of solutions

A|B matrix will be given as -

$\begin{bmatrix} 1 &2 &0 &|5 \\ 4& 8 &0 & |12 \\ 3& 6 & 3 & |15 \end{bmatrix}$

After the row-operations -   R3->R3-3R1  and

R2->4R1-R2

We obtain the matrix -

$\begin{bmatrix} 1 &2 &0 &|5 \\ 0 & 0& 0& |8\\ 0& 0 & 0 & |0 \end{bmatrix}$

Now since, $Rank(A)\neq Rank(A|B)$

Hence, this is an inconsistent solution, that is, no solution.

There are no solutions.

If we multiply 1st equation by 4, we get

4x + 8y = 20

But 2nd equation says

4x + 8y = 12

Clearly, there can not be any pair of (x,y), which satisfies both equations.
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I am getting rank of augmented matrix as 3 as well as rank of A =3 so according to this condition there must be a unique solution , although it shouldn't exist but then how can it contradict ?
Rank of A is not 3. Recheck your method.

The rank of augemented matrix(|AB|=3) and coefficient matrix|A|=2 is not same! hence there is no solution.