Solving a linear system of equations use row operations Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?
$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$
Every time I solve this it feels like it could have been done simpler... Thanks!

My attempt:
I first exchanged row 1 and row 2  to get:
$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$ $$3x_1-2x_2+3x_3=8$$
Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:
$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$
But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!

We are given
$$\begin{pmatrix}
2 & 4 & -1\\
1 & 1 & -1\\
3 & -2 & 3
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
0\\
8
\end{pmatrix}$$
Take $R_1 - 2R_2 \to R_2$ and $-3R_1+2R_3 \to R_3$ to form
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & -16 & 9
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
-5
\end{pmatrix}$$
Then let $8R_2 + R_3 \to R_3$ to obtain
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & 0 & 17
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
51
\end{pmatrix}$$
We then find that $x_1=1,x_2=2,x_3=3$.
