Proving type of solutions for matrix

So the question is:

Consider a system of linear equations written in matrix form as Ax =b, where A is a m×n–

matrix, x is an n–dimensional vector–column, and b is an m–dimensional vector–column.

Show that the system has

(a) no solutions if and only if rank A < rank [A|b], where [A|b] is the augmented matrix;

(b) a unique solution if and only if rank A = rank [A|b] = n;

(c) infinitely many solutions if and only if rank A = rank [A|b] < n.

I think I more or less know what Im doing but im confused by one thing.

What I first did was assume that the matrix was in reduced row echelon form and took it from there.

For parts B and C could the x value be zero for a given variable but have the b value be a non-zero number. Wouldnt this still maintain that the rank was equal but there be no solutions. Ill try and explain with an example:

A given row in matrix A is

1 0 0 0 0

X is given by [0,x,x,x,x]

b for this particular row is a non zero value.

Wouldnt this still maintain the rank however make it so there is no solution?

Also is there a way to solve this problem mathematically? My proof has pretty much been all words which I dont find to be a very concrete answer.

Thanks