Answer
$(-2,5,-2)$
Work Step by Step
To solve the system $\begin{cases}x+y \hspace{1.5cm}=3 \\ \hspace{9mm} 2y \hspace{13mm}=10 \\ 3x+2y-4z=12 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal (if possible).
The corresponding augmented matrix is
$$\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\
0 & 2 & 0 & 10\\
3 & 2 & -4 & 12\\
\end{array}
\right].$$
We replace Row_3 with Row_3-3*Row_1 to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\
0 & 2 & 0 & 10\\
0 & -1 & -4 & 3\\
\end{array}
\right].$$
Next, we multiply Row_2 by $\dfrac{1}{2}$ to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\
0 & 1 & 0 & 5\\
0 & -1 & -4 & 3\\
\end{array}
\right].$$
Thirdly, we replace Row_3 with Row_3+Row_2 to obtain the equaivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\
0 & 1 & 0 & 5\\
0 & 0 & -4 & 8\\
\end{array}
\right].$$
Fourthly, we multiply Row_3 by $-\dfrac{1}{4}$to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\
0 & 1 & 0 & 5\\
0 & 0 & 1 & -2\\
\end{array}
\right].$$
Now, we form the system of equations corresponding to this matrix:
$$\begin{cases}x+y \hspace{10mm}=3 \\ \hspace{9mm} y \hspace{8mm}=5 \\ \hspace{15mm} z=-2 \\ \end{cases}.$$
We see $z=-2$ and $y=5$, and to find our value for $x$, we plug $y=5$ into the first equation and solve it for $x$.
Thus
$$x+5=3 \\ x=-2.$$
Thus we have exactly one solution, which is $(-2,5,-2).$
