Answer
$x_1=-4t$, $x_2=-\frac{1}{2}t$, $x_3=t$
Work Step by Step
The augmented matrix is given by:
$\left[ {\begin{array}{*{20}{c}}
2&{ - 8}&4&0\\
3&{ - 10}&7&0\\
0&{10}&5&0
\end{array}} \right]$
Row reducing this matrix gives
$\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
2&{ - 8}&4&0\\
3&{ - 10}&7&0\\
0&{10}&5&0
\end{array}} \right]\\
{R_1} \to \frac{{{R_1}}}{2}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 4}&2&0\\
3&{ - 10}&7&0\\
0&{10}&5&0
\end{array}} \right]\\
{R_2} \to {R_2} - 3{R_1}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 4}&2&0\\
0&2&1&0\\
0&{10}&5&0
\end{array}} \right]\\
{R_3} \to {R_3} - 5{R_2}\\
\left[ {\begin{array}{*{20}{c}}
1&{ - 4}&2&0\\
0&2&1&0\\
0&0&0&0
\end{array}} \right]\\
{R_1} \to {R_1} + 2{R_2}\\
\left[ {\begin{array}{*{20}{c}}
1&0&4&0\\
0&2&1&0\\
0&0&0&0
\end{array}} \right]\\
{R_2} \to \frac{{{R_2}}}{2}\\
\left[ {\begin{array}{*{20}{c}}
1&0&4&0\\
0&1&{1/2}&0\\
0&0&0&0
\end{array}} \right]
\end{array}$
This means that $x_3$ is a free variable.
Let $x_3=t$; then $x_1=-4x_3=-4t$, $x_2=-\frac{1}{2}x_3=-\frac{1}{2}t$
