Answer
$(1,2,4)$
Work Step by Step
To solve the system $\begin{cases}5x \hspace{15mm}=5 \\ 2x+y \hspace{5mm}=4 \\ 3x+y-5z=-15 \\ \end{cases},$ 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
$$\left[
\begin{array}{ccc|c}
5 & 0 & 0 & 5 \\
2 & 1 & 0 & 4\\
3 & 1 & -5 & -15\\
\end{array}
\right].$$
First, we multiply Row_1 by $\dfrac{1}{5}$ to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
2 & 1 & 0 & 4\\
3 & 1 & -5 & -15\\
\end{array}
\right].$$
Next, we replace Row_2 with Row_2-2Row_1 to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 2\\
3 & 1 & -5 & -15\\
\end{array}
\right].$$
Thirdly, we replace Row_3 with Row_3-3Row_1 to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 2\\
0 & 1 & -5 & -18\\
\end{array}
\right].$$
Fourthly, we replace Row_3 with Row_3-Row_2 to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 2\\
0 & 0 & -5 & -20\\
\end{array}
\right].$$
Fifthly, we multiply Row_3 by $-\dfrac{1}{5}$ to obtain the equivalent matrix
$$\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 4\\
\end{array}
\right].$$
Now, we form the system of equations corresponding to this matrix:
$$\begin{cases}x \hspace{15mm} =1 \\ \hspace{5mm} y \hspace{9mm}=2 \\ \hspace{10mm} z=4 \\ \end{cases}.$$
We see $x=1$, $y=2$, and $z=4$.
Thus we have exactly one solution, which is $(1,2,4).$