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