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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

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.