= Solution
Apply the same <elementary row operations> to the <augmented matrix>:
$$
\left(\begin{array}{rrrr|r}1&2&-1&4&2\\1&2&4&1&7\\-1&-2&6&-7&3\end{array}\right)\sim\left(\begin{array}{rrrr|r}1&2&-1&4&2\\0&0&5&-3&5\\0&0&0&0&0\end{array}\right).
$$
The last row establishes consistency; equivalently the third original row is the second minus twice the first, and the right-hand entries obey $3=7-2\cdot2$. With $x_2=s$ and $x_4=5t$, the second equation gives $x_3=1+3t$ and the first gives $x_1=3-2s-17t$. All real solutions are therefore
$$
\boxed{x=\begin{pmatrix}3\\0\\1\\0\end{pmatrix}+s\begin{pmatrix}-2\\1\\0\\0\end{pmatrix}+t\begin{pmatrix}-17\\0\\3\\5\end{pmatrix},\qquad s,t\in\mathbb R.}
$$
This <affine solution space> is one particular solution plus the <null space>. The two <free variables> show that there are no additional restrictions.
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