Solve the system of equations by Gauss Elimination method
3x+4y+5=18; 2x-y+8=13; 5x-2y+7z=20.

Let's solve the system of equations using the Gauss Elimination method. The given system of equations is:

\[
\begin{aligned}
1) & \ 3x + 4y + 5z = 18 \\
2) & \ 2x - y + 8z = 13 \\
3) & \ 5x - 2y + 7z = 20
\end{aligned}
\]

### Step 1: Write the augmented matrix
The first step is to write the system of equations in an augmented matrix form:

\[
\begin{pmatrix}
3 & 4 & 5 & | & 18 \\
2 & -1 & 8 & | & 13 \\
5 & -2 & 7 & | & 20
\end{pmatrix}
\]

### Step 2: Apply row operations to obtain the row echelon form

**Row 1**: No changes (we'll start with this as the pivot row).

\[
R1 = \begin{pmatrix} 3 & 4 & 5 & | & 18 \end{pmatrix}
\]

**Row 2**: Eliminate the first element in Row 2 by subtracting \(\frac{2}{3}\) of Row 1 from Row 2:

\[
R2 \rightarrow R2 - \frac{2}{3}R1
\]

\[
R2 = \begin{pmatrix} 2 & -1 & 8 & | & 13 \end{pmatrix} - \frac{2}{3} \times \begin{pmatrix} 3 & 4 & 5 & | & 18 \end{pmatrix}
\]

\[
R2 = \begin{pmatrix} 0 & -\frac{11}{3} & \frac{14}{3} & | & 7 \end{pmatrix}
\]

**Row 3**: Eliminate the first element in Row 3 by subtracting \(\frac{5}{3}\) of Row 1 from Row 3:

\[
R3 \rightarrow R3 - \frac{5}{3}R1
\]

\[
R3 = \begin{pmatrix} 5 & -2 & 7 & | & 20 \end{pmatrix} - \frac{5}{3} \times \begin{pmatrix} 3 & 4 & 5 & | & 18 \end{pmatrix}
\]

\[
R3 = \begin{pmatrix} 0 & -\frac{22}{3} & \frac{2}{3} & | & -10 \end{pmatrix}
\]

So, the matrix now looks like:

\[
\begin{pmatrix}
3 & 4 & 5 & | & 18 \\
0 & -\frac{11}{3} & \frac{14}{3} & | & 7 \\
0 & -\frac{22}{3} & \frac{2}{3} & | & -10
\end{pmatrix}
\]

### Step 3: Further elimination to make it upper triangular

**Row 3**: Eliminate the second element in Row 3 by subtracting \(\frac{2}{11}\) of Row 2 from Row 3:

\[
R3 \rightarrow R3 - \frac{2}{11}R2
\]

\[
R3 = \begin{pmatrix} 0 & -\frac{22}{3} & \frac{2}{3} & | & -10 \end{pmatrix} - \frac{2}{11} \times \begin{pmatrix} 0 & -\frac{11}{3} & \frac{14}{3} & | & 7 \end{pmatrix}
\]

\[
R3 = \begin{pmatrix} 0 & 0 & -4 & | & -\frac{48}{11} \end{pmatrix}
\]

### Step 4: Back-substitution

Now that we have the matrix in upper triangular form, we can back-substitute to find the solutions:

**From Row 3**:
\[
-4z = -\frac{48}{11} \implies z = \frac{12}{11}
\]

**From Row 2**:
\[
-\frac{11}{3}y + \frac{14}{3}\left(\frac{12}{11}\right) = 7
\]
\[
-\frac{11}{3}y + \frac{168}{33} = 7
\]
\[
-\frac{11}{3}y = 7 - \frac{168}{33} \implies y = \frac{5}{11}
\]

**From Row 1**:
\[
3x + 4\left(\frac{5}{11}\right) + 5\left(\frac{12}{11}\right) = 18
\]
\[
3x + \frac{20}{11} + \frac{60}{11} = 18
\]
\[
3x = 18 - \frac{80}{11} \implies x = \frac{34}{11}
\]

### Final Solution:
\[
x = \frac{34}{11}, \quad y = \frac{5}{11}, \quad z = \frac{12}{11}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How is the Gauss-Jordan elimination method different from the Gauss elimination method?
2. What are the potential pitfalls when performing row operations manually?
3. Can a system of linear equations have no solution, and how does Gauss elimination handle that?
4. What is the significance of the determinant in solving a system of linear equations?
5. How does Gauss elimination relate to matrix inversion?

**Tip:** Always check the consistency of the system before starting with Gauss elimination to determine if a solution exists or if the system is inconsistent.

Learn how to solve a system of equations using the Gauss Elimination method. This problem involves linear algebra and systems of linear equations, suitable for high school students.

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