Answer
( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $\ \ t\in \mathbb{R}$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
1 & -2 & 5 & 3\\
-2 & 6 & -11 & 1\\
3 & -16 & 20 & -26
\end{array}\right] \ \ \begin{array}{l}
.\\
+2R_{2}.\\
-3R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & -2 & 5 & 3\\
0 & 2 & -1 & 7\\
0 & -10 & 5 & -35
\end{array}\right] \ \ \begin{array}{l}
+R_{2}.\\
\div 2.\\
-5R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & 4 & 10\\
0 & 1 & -1/2 & 7/2\\
0 & 0 & 0 & 0
\end{array}\right] \ \ \begin{array}{l}
.\\
.\\
.
\end{array}$
Last row has all zeros, so
the system is dependent (infinite number of solutions).
Taking $w=t\in \mathbb{R}$ (arbitrary),
back-substituting:
$x=10-4t$
$y=\displaystyle \frac{7}{2}+\frac{1}{2}t$
Solution set: $\{$( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $t\in \mathbb{R} \}$
