Answer
$x_1-x_2+3x_4+x_5=2, x_1+x_2+2x_3+x_4-x_5=4, x_2+2x_4+3x_5=0$
Work Step by Step
The augmented matrix for a system of two linear equations ($m$ and $n$) can be written as:
$a_{11}x_1+.....+a_{1n}x_n=b_1 \\ \vdots
\\a_{m1}x_1+.....+a_{mn}x_n=b_m$
which is equivalent to: $\left[\begin{array}{ccc|c} a_{11} & \cdots & a_{1n} &b_1\\a_{21} & \cdots & a_{2n} &b_2\\ \vdots & \vdots & \vdots & \vdots \\a_{m1} &\cdots &a_{mn} & b_m \end{array}\right]$
Now, from the given augmented matrix, the corresponding system of linear equations are:
$x_1-x_2+3x_4+x_5=2, x_1+x_2+2x_3+x_4-x_5=4, x_2+2x_4+3x_5=0$
