System of Equation 
Solve the system of equations
$$-x_1+x_2+x_3=a$$ $$x_1-x_2+x_3=b$$ $$x_1+x_2-x_3=c$$

I have tried writing the augmented matrix of the system of equations above and reducing it into echelon form but that didn't work out. Please help.
 A: Adding the first two equations, $$2x_{3}=a+b$$
Can you proceed in the similar manner?
A: Let's do the Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
-1 & 1 & 1 & a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&& R_1\gets -R_1
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 0 & 2 & b+a \\
0 & 2 & 0 & c+a
\end{array}\right]
&& \begin{aligned}R_2&\gets R_2-R_1\\R_3&\gets R_3-R_1\end{aligned}
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 2 & 0 & c+a \\
0 & 0 & 2 & b+a 
\end{array}\right]
&& R_2\leftrightarrow R_3
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&& \begin{aligned}R_2&\gets \tfrac{1}{2}R_2\\R_3&\gets \tfrac{1}{2}R_3\end{aligned}
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & 0 & (b-a)/2 \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&&R_1\gets R_1+R_3
\\
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 0 & (b+c)/2 \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&&R_1\gets R_1+R_2
\end{align}
A: For a $3\times 3$ system with only $1$ and $-1$ coefficients, it's easier to isolate one unknown from one equation and to replace it in an other equation. Here for example equation $3$ gives you : $x_3 = x_1 + x_2 - c$. Then you can replace $x_3$ in equation $2$, isolate $x_2$ and replace in eq. $1$. 
