Answer
$ (\displaystyle \frac{10}{3},\frac{1}{3})$
Work Step by Step
Write the augmented matrix and,
using row transformations arrive at the reduced row echelon form.
-----------
Swap equations 1 and 2,
remove decimals/fractions
$\left[\begin{array}{lll}
0.1 & -0.1 & 0.3\\
0.5 & 0.1 & 1.7\\
1 & 1 & 11/3
\end{array}\right]\begin{array}{l}
10R_{1}.\\
10R_{2}.\\
3R_{3}.
\end{array}$
$\left[\begin{array}{lll}
1 & - 1 & 3\\
5 & 1 & 17\\
3 & 3 & 11
\end{array}\right]\begin{array}{l}
.\\
R_{2}-5R_{1}.\\
R_{3}-3R_{1}.
\end{array}$
... clears column $1,$
$\left[\begin{array}{lll}
1 & - 1 & 3\\
0 & 6 & 2\\
0 & 6 & 2
\end{array}\right]\begin{array}{l}
.\\
.\\
R_{3}-R_{2}.
\end{array}$
... begin clearing column 2,
$\left[\begin{array}{lll}
1 & - 1 & 3\\
0 & 6 & 2\\
0 & 0 & 0
\end{array}\right]\begin{array}{l}
.\\
\frac{1}{6}R_{2}\\
.
\end{array}$
... leading 1 in row 2,
$\left[\begin{array}{lll}
1 & - 1 & 3\\
0 & 1 & 1/3\\
0 & 0 & 0
\end{array}\right]\begin{array}{l}
R_{1}+R_{2}.\\
.\\
.
\end{array}$
... clears column 2,
$\left[\begin{array}{lll}
1 & 0 & 10/3\\
0 & 1 & 1/3\\
0 & 0 & 0
\end{array}\right]$
Last row: all zeros, so system is consistent.
$x=10/3$
$y=1/3$
Solution: $ (\displaystyle \frac{10}{3},\frac{1}{3})$
