Answer
$(-4,2)$
Work Step by Step
To solve the system $\begin{cases}x+3y=2 \\ x+2y=0 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal.
The corresponding augmented matrix is
$$\left[
\begin{array}{cc|c}
1 & 3 & 2 \\
1 & 2 & 0\\
\end{array}
\right].$$
Next, we replace Row_2 with Row_2-Row_1 to obtain the equivalent matrix
$$\left[
\begin{array}{cc|c}
1 & 3 & 2 \\
0 & -1 & -2\\
\end{array}
\right].$$
Next, we multiply Row_2 by $-1$ to obtain the equaivalent matrix
$$\left[
\begin{array}{cc|c}
1 & 3 & 2 \\
0 & 1 & 2\\
\end{array}
\right].$$
Now, we form the system of equations corresponding to this matrix
$$\begin{cases}x+3y=2 \\ y=2 \\ \end{cases}.$$
We see $y=2$, and to find our value for $x$, we plug $y=2$ into the first equation and solve it for $x$.
Thus
$$x+3(2)=2 \\ x+6=2\\ x=-4.$$
Thus we have exactly one solution, which is $(-4,2).$