What would the solution be if I cannot simplify my solution to find more basic variables(pivots)? 4x3 matrix $$x - 5y + 4z = -3$$
$$2x - 7y + 3z = -2$$
$$-2x + y +  7z = -1$$
The furthest I was able to reach before coming to the conclusion that I could not move any further was:
Row 1 => $x - 5y + 4z = -3$
Row 2 => $9y + 15z = -7$
Row 3 => $8y + 5z = 1$
Please explain to me how I would be able to express my solution in the simplest form! Also let me know what type of solution this system would contain (independent, dependent, inconsistent)

Here is the reduction of the augmented matrix.
Column 1
$$
  \left[
\begin{array}{rcc}
 1 & 0 & 0 \\
 -2 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{rrc|ccc}
 1 & -5 & 4 & 1 & 0 & 0 \\
 2 & -7 & 3 & 0 & 1 & 0 \\
 -2 & 1 & 7 & 0 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & -5 & 4 & 1 & 0 & 0 \\
 0 & 3 & -5 & -2 & 1 & 0 \\
 0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
$$
Column 2
$$
  \left[
\begin{array}{ccc}
 1 & \frac{5}{3} & 0 \\
 0 & \frac{1}{3} & 0 \\
 0 & 3 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & -5 & 4 & 1 & 0 & 0 \\
 0 & 3 & -5 & -2 & 1 & 0 \\
 0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
 \boxed{1} & 0 & -\frac{13}{3} & -\frac{7}{3} & \frac{5}{3} & 0 \\
 0 & \boxed{1} & -\frac{5}{3} & -\frac{2}{3} & \frac{1}{3} & 0 \\
 0 & 0 & 0 & -4 & 3 & 1 \\
\end{array}
\right]
$$
Least squares solution
The least squares solution to $$\mathbf{A}x = b$$ is 
$$
 x = \frac{1}{5827}
%
 \left[
\begin{array}{r}
 -257 \\ 1369 \\ -1168 \\
\end{array}
\right]
$$
The least squares residual errors is
$$
\begin{align}
  \mathbf{A} x - b &= \mathbf{0} \\
%
 \left[
\begin{array}{rrc}
 1 & -5 & 4 \\
 2 & -7 & 3 \\
 -2 & 1 & 7 \\
\end{array}
\right]
\frac{1}{5827}
%
 \left[
\begin{array}{r}
 -257 \\ 1369 \\ -1168 \\
\end{array}
\right]
%
 - 
 \left[
\begin{array}{r}
 -3 \\ -2 \\ -1 \\
\end{array}
\right]
%
&=
\frac{1}{26}
 \left[
\begin{array}{r}
 20 \\ -15 \\ -5 \\
\end{array}
\right]
%
\end{align}
$$
