Write the solution of the linear system corresponding to the reduced augmented matrix.
\[
\left[\begin{array}{rrr|r}
1 & 0 & -8 & 5 \\
0 & 1 & 2 & -7
\end{array}\right]
\]
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A. The unique solution is $x_{1}=\quad, x_{2}=\ldots$, and $x_{3}=$ (Simplify your answers.)
B. The system has infinitely many solutions. The solution is $x_{1}=\left[\quad, x_{2}=\left[\right.\right.$, and $x_{3}=t$. (Simplify your answers. Type expressions using t as the variable.)
C. There is no solution.

Step 1 ：The given matrix is in reduced row echelon form. This form directly gives us the solution to the system of equations. The first row corresponds to the equation \(x_1 - 8x_3 = 5\) and the second row corresponds to the equation \(x_2 + 2x_3 = -7\). We can solve these equations directly for \(x_1\) and \(x_2\) in terms of \(x_3\).

Step 2 ：From the first equation, we can express \(x_1\) in terms of \(x_3\) as \(x_1 = 8x_3 + 5\).

Step 3 ：From the second equation, we can express \(x_2\) in terms of \(x_3\) as \(x_2 = -2x_3 - 7\).

Step 4 ：Thus, the unique solution is \(x_{1}=8x_{3} + 5\), \(x_{2}=-2x_{3} - 7\), and \(x_{3}=x_{3}\). This corresponds to the system having infinitely many solutions. The solution is \(x_{1}=8x_{3} + 5\), \(x_{2}=-2x_{3} - 7\), and \(x_{3}=t\).

Step 5 ：\(\boxed{x_{1}=8x_{3} + 5, x_{2}=-2x_{3} - 7, x_{3}=t}\) is the final answer.