Answer
$(\displaystyle \frac{1}{5},\frac{4}{5})$
Work Step by Step
Write the augmented matrix and,
using row transformations arrive at the reduced row echelon form (rref).
-----------
$\left[\begin{array}{lll}
1 & 1 & 1\\
3 & -2 & -1\\
5 & -1 & 1/5
\end{array}\right]\begin{array}{l}
.\\
R_{2}-3R_{1}.\\
R_{3}-5R_{1}.
\end{array}$, clear column $1$
$\left[\begin{array}{lll}
1 & 1 & 1\\
0 & -5 & -4\\
0 & -6 & -24/5
\end{array}\right]\begin{array}{l}
.\\
.\\
5R_{3}-6R_{2}.
\end{array}$, clear column 2$...$
$\left[\begin{array}{lll}
1 & 1 & 1\\
0 & -5 & -4\\
0 & 0 & 0
\end{array}\right]\begin{array}{l}
.\\
-\frac{1}{5}R_{1}\\
.
\end{array}$ , leading 1 in row 2
$\left[\begin{array}{lll}
1 & 1 & 1\\
0 & 1 & 4/5\\
0 & 0 & 0
\end{array}\right]\begin{array}{l}
R_{1}-R_{2}.\\
.\\
.
\end{array}$, clear column 2
$\left[\begin{array}{lll}
1 & 0 & 1/5\\
0 & 1 & 4/5\\
0 & 0 & 0
\end{array}\right]$
System is consistent (last row has all zeros)
$x=1/5$
$y=4/5$
Solution: $(\displaystyle \frac{1}{5},\frac{4}{5})$
