= Solution
The second, third and fourth <linear equations> give
$$
x_1+x_2=2,\qquad x_2+x_3=0,\qquad x_3+x_4=-1.
$$
Setting $x_4=t$ yields $x_3=-1-t$, $x_2=1+t$ and $x_1=1-t$. The first <linear equation> then gives $(1-t)+(1+t)+(-1-t)+t=1$, and the fifth gives $(1-t)+t=1$, so both remaining conditions hold for every real $t$. Consequently
$$
\boxed{\mathbf x=\begin{pmatrix}1\\1\\-1\\0\end{pmatrix}
+t\begin{pmatrix}-1\\1\\-1\\1\end{pmatrix},\qquad t\in\mathbb R.}
$$
This is a particular solution plus the whole <null space>, an <affine line in a vector space> of solutions. The derivation both produces solutions and shows that none have been omitted.
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