Gaussian elimination with $4x-3y=11$ and $x-3y=-1$

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Excuse syntax.

I would like to use Gaussian elimination to solve the system $[4x+3y=11]$ and $[x-3y=-1]$

I'm following the exercise 'Try it #3' on this online text book page: enter image description here

The solution provided is over here: enter image description here

The solution is $(2, 1)$ whereas I got $(0, -3.67)$

Since I don't know the syntax for multiple rows within a square brace [] I'm going to screen shot my working:

enter image description here

ON the left column I work towards having all 1 in the diagonal and 0 underneath this diagonal line.

First I swap the rows.

Then I multiple new row 1 by -4 and add it to row 2 to delete the 4.

Then I divide row 2 by 12.

This gives me $y=-3.67$ and I use this on the right column to back substitute and solve to get the $(x,y)$ points.

I got $(0, -3.67)$ whereas the right answer is $(2,1)$. Where did I go wrong and how can I arrive at $(2,1)$?

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We will use Gaussian Elimination. Our augmented matrix is $$ \left[\begin{array}{rr|r} 4 & 3 & 11 \\ 1 & -3 & -1 \end{array}\right] $$

$R_1 \leftrightarrow R_2$

$$ \left[\begin{array}{rr|r} 1 & -3 & -1 \\ 4 & 3 & 11 \end{array}\right] $$

$R_2 \rightarrow R_2 - 4 R_1$

$$ \left[\begin{array}{rr|r} 1 & -3 & -1 \\ 0 & 15 & 15 \end{array}\right] $$

$R_2 \rightarrow R_2/15$

$$ \left[\begin{array}{rr|r} 1 & -3 & -1 \\ 0 & 1 & 1 \end{array}\right] $$

$R_1 \rightarrow R_1 + 3 R_2$

$$ \left[\begin{array}{rr|r} 1 & 0 & 2 \\ 0 & 1 & 1 \end{array}\right] $$

Reading from the bottom row up, we have

$$y = 1, x = 2$$