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To solve this system using Cramer's Rule, let’s first write down the equations clearly:

1. \( x + 3y + z = -1 \)
2. \( x + 2y + 3z = 4 \)
3. \( x + 8y + 5z = -7 \)

Cramer's Rule uses determinants to solve systems of linear equations of the form:

\[
Ax = B
\]

where \( A \) is the matrix of coefficients, \( x \) is the vector of variables, and \( B \) is the constants vector.

### Step 1: Matrix Setup
Let’s define the coefficient matrix \( A \) and the constant vector \( B \):

\[
A = \begin{bmatrix} 1 & 3 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -1 \\ 4 \\ -7 \end{bmatrix}
\]

### Step 2: Determinant of \( A \)
To find \( x \), \( y \), and \( z \) using Cramer's Rule, we first calculate \( \det(A) \).

\[
\det(A) = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 5 \end{vmatrix}
\]

Expanding along the first row:

\[
\det(A) = 1 \cdot \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 8 \end{vmatrix}
\]

Calculating each minor:

1. \( \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} = (2 \cdot 5) - (3 \cdot 8) = 10 - 24 = -14 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & 5 \end{vmatrix} = (1 \cdot 5) - (3 \cdot 1) = 5 - 3 = 2 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 8 \end{vmatrix} = (1 \cdot 8) - (2 \cdot 1) = 8 - 2 = 6 \)

So,

\[
\det(A) = 1 \cdot (-14) - 3 \cdot 2 + 1 \cdot 6 = -14 - 6 + 6 = -14
\]

### Step 3: Determinants for \( x \), \( y \), and \( z \)

Now, we replace each column of \( A \) with \( B \) to form matrices \( A_x \), \( A_y \), and \( A_z \) and compute their determinants.

#### \( \det(A_x) \):

\[
A_x = \begin{bmatrix} -1 & 3 & 1 \\ 4 & 2 & 3 \\ -7 & 8 & 5 \end{bmatrix}
\]

Expanding along the first row:

\[
\det(A_x) = -1 \cdot \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} - 3 \cdot \begin{vmatrix} 4 & 3 \\ -7 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 2 \\ -7 & 8 \end{vmatrix}
\]

Using the minors computed:

1. \( \begin{vmatrix} 2 & 3 \\ 8 & 5 \end{vmatrix} = -14 \)
2. \( \begin{vmatrix} 4 & 3 \\ -7 & 5 \end{vmatrix} = (4 \cdot 5) - (3 \cdot -7) = 20 + 21 = 41 \)
3. \( \begin{vmatrix} 4 & 2 \\ -7 & 8 \end{vmatrix} = (4 \cdot 8) - (2 \cdot -7) = 32 + 14 = 46 \)

Thus,

\[
\det(A_x) = (-1)(-14) - 3(41) + 1(46) = 14 - 123 + 46 = -63
\]

#### \( \det(A_y) \):

\[
A_y = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 4 & 3 \\ 1 & -7 & 5 \end{bmatrix}
\]

Expanding and solving similarly, we find \( \det(A_y) \).

#### \( \det(A_z) \):

\[
A_z = \begin{bmatrix}

Learn how to solve a system of three linear equations using Cramer's Rule, including determinant calculations and substitutions. This problem demonstrates concepts in linear algebra, suitable for high school students in Grades 10-12.

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