**[Part 3 of 3]** Solve the system of equation given in part 1 by performing back substitution on the echelon form of the augmented matrix determined in part 2.

To solve the system of equations using back substitution, we will use the row-echelon form of the augmented matrix obtained earlier:

\[
\begin{bmatrix}
1 & 1 & 4 & -11 \\
0 & 1 & \frac{19}{7} & -\frac{60}{7} \\
0 & 0 & 1 & -10
\end{bmatrix}
\]

This corresponds to the following system of equations:

1. \( x_1 + x_2 + 4x_3 = -11 \)
2. \( x_2 + \frac{19}{7}x_3 = -\frac{60}{7} \)
3. \( x_3 = -10 \)

### Step 1: Solve for \( x_3 \) using the third equation.
From the third equation:
\[
x_3 = -10
\]

### Step 2: Substitute \( x_3 = -10 \) into the second equation to solve for \( x_2 \).
The second equation is:
\[
x_2 + \frac{19}{7}x_3 = -\frac{60}{7}
\]

Substituting \( x_3 = -10 \):
\[
x_2 + \frac{19}{7}(-10) = -\frac{60}{7}
\]

Simplifying:
\[
x_2 - \frac{190}{7} = -\frac{60}{7}
\]

Adding \( \frac{190}{7} \) to both sides:
\[
x_2 = \frac{190}{7} - \frac{60}{7} = \frac{130}{7}
\]

So, \( x_2 = \frac{130}{7} \).

### Step 3: Substitute \( x_2 = \frac{130}{7} \) and \( x_3 = -10 \) into the first equation to solve for \( x_1 \).
The first equation is:
\[
x_1 + x_2 + 4x_3 = -11
\]

Substituting the values of \( x_2 \) and \( x_3 \):
\[
x_1 + \frac{130}{7} + 4(-10) = -11
\]

Simplifying:
\[
x_1 + \frac{130}{7} - 40 = -11
\]

First, express \( -40 \) as a fraction with a denominator of 7:
\[
x_1 + \frac{130}{7} - \frac{280}{7} = -11
\]

Combine the fractions:
\[
x_1 - \frac{150}{7} = -11
\]

Express \( -11 \) as a fraction with a denominator of 7:
\[
x_1 - \frac{150}{7} = -\frac{77}{7}
\]

Adding \( \frac{150}{7} \) to both sides:
\[
x_1 = \frac{150}{7} - \frac{77}{7} = \frac{73}{7}
\]

### Final Solution:
The solution to the system of equations is:
\[
x_1 = \frac{73}{7}, \quad x_2 = \frac{130}{7}, \quad x_3 = -10
\]

Would you like further details or have any questions? Here are five related questions to explore:

1. How do you identify dependent and independent variables in a system of equations?
2. How can you determine if a system has no solution based on its row-echelon form?
3. How would the solution change if one equation was inconsistent (e.g., \( 0 = 1 \))?
4. What does it mean geometrically when a system of three equations is consistent?
5. How does back substitution differ when working with matrices in reduced row-echelon form?

**Tip:** Always start back substitution from the last row and work upwards for an efficient solution process.

Learn how to solve a system of linear equations using back substitution method. Step-by-step solution provided using row-echelon form of the augmented matrix. Suitable for high school students studying algebra and linear systems.

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