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To start, we can find the $y$-intercept of each of these lines. Using this, we can calculate the length of that side of the triangle, and use it as a base. Letting $x=0$ in the first equation gives $y=-2$ as a $y$-intercept. Letting $x=0$ in the second equation gives $3y=12\Rightarrow y=4$ as a $y$-intercept. Therefore, the triangle has a length of $4-(-2)=6$ on the $y$-axis.

The height of the triangle will be equal to the $x$-coordinate of the intersection of the two lines. So, we need to solve for $x$ in the system: $$\begin{array}{rl}y-3x& =-2\\ 3y+x& =12\end{array}$$Multiply the first equation by 3, then subtract the second equation as shown: \begin{array}{ r c c c l} $3y%%DISP_1%%amp;-&$9x%%DISP_1%%amp;=&-6\\ -($3y%%DISP_1%%amp;+&$x%%DISP_1%%amp;=&12)\\ \hline &-&$10x%%DISP_1%%amp;=&-18\\ \end{array}Therefore, $x=\frac{18}{10}=\frac{9}{5}$. This is equal to the height of the triangle. The area will be $\frac{1}{2}\cdot \frac{9}{5}\cdot 6=\boxed{\frac{27}{5}}$