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To solve this system, let's start by writing down the equations as given and then analyze the system of linear equations. The equations provided are:

\[
\begin{aligned}
1.\ & x_1 - 5x_2 + 2x_3 - 3x_5 - 3x_6 = -1 \\
2.\ & x_1 - 5x_2 = 0 \\
3.\ & -x_4 - 3x_5 - 2x_6 = -3 \\
4.\ & -7x_5 - 5x_6 = -3
\end{aligned}
\]

### Step 1: Solve the simpler equations
From the second equation:
\[
x_1 - 5x_2 = 0 \quad \Rightarrow \quad x_1 = 5x_2
\]

From the fourth equation:
\[
-7x_5 - 5x_6 = -3 \quad \Rightarrow \quad 7x_5 + 5x_6 = 3
\]

This can be solved for \( x_5 \) in terms of \( x_6 \):
\[
x_5 = \frac{3 - 5x_6}{7}
\]

### Step 2: Substitute into the remaining equations
Substitute \( x_1 = 5x_2 \) into the first equation:
\[
5x_2 - 5x_2 + 2x_3 - 3x_5 - 3x_6 = -1 \quad \Rightarrow \quad 2x_3 - 3x_5 - 3x_6 = -1
\]

Substitute \( x_5 = \frac{3 - 5x_6}{7} \) into the above equation:
\[
2x_3 - 3\left(\frac{3 - 5x_6}{7}\right) - 3x_6 = -1
\]
Multiply through by 7 to eliminate the fraction:
\[
14x_3 - 9 + 15x_6 - 21x_6 = -7 \quad \Rightarrow \quad 14x_3 - 6x_6 = 2
\]
This simplifies to:
\[
7x_3 - 3x_6 = 1 \quad \Rightarrow \quad x_3 = \frac{1 + 3x_6}{7}
\]

### Step 3: Solve for the remaining variables
Now use the third equation:
\[
-x_4 - 3x_5 - 2x_6 = -3
\]
Substitute \( x_5 = \frac{3 - 5x_6}{7} \):
\[
-x_4 - 3\left(\frac{3 - 5x_6}{7}\right) - 2x_6 = -3
\]
Multiply through by 7:
\[
-7x_4 - 9 + 15x_6 - 14x_6 = -21 \quad \Rightarrow \quad -7x_4 + x_6 = -12
\]
So:
\[
x_4 = \frac{12 + x_6}{7}
\]

### Step 4: Express the solution
We can now write the general solution to the system. The variables \( x_2 \), \( x_6 \), and another free variable (let's denote it as \( u \)) are free parameters. The solution can be expressed as:

\[
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6
\end{pmatrix} = 
\begin{pmatrix}
5x_2 \\
x_2 \\
\frac{1 + 3x_6}{7} \\
\frac{12 + x_6}{7} \\
\frac{3 - 5x_6}{7} \\
x_6
\end{pmatrix} = 
\begin{pmatrix}
0 \\
0 \\
\frac{1}{7} \\
\frac{12}{7} \\
\frac{3}{7} \\
0
\end{pmatrix} + 
x_2
\begin{pmatrix}
5 \\
1 \\
0 \\
0 \\
0 \\
0
\end{pmatrix} + 
x_6
\begin{pmatrix}
0 \\
0 \\
\frac{3}{7} \\
\frac{1}{7} \\
-\frac{5}{7} \\
1
\end{pmatrix}
\]

So the general solution is a linear combination of these vectors, plus a particular solution vector. 

### Summary:
The solution is:
\[
\mathbf{x} = 
\begin{pmatrix}
0 \\
0 \\
\frac{1}{7} \\
\frac{12}{7} \\
\frac{3}{7} \\
0
\end{pmatrix} 
+ s
\begin{pmatrix}
5 \\
1 \\
0 \\
0 \\
0 \\
0
\end{pmatrix}
+ t
\begin{pmatrix}
0 \\
0 \\
\frac{3}{7} \\
\frac{1}{7} \\
-\frac{5}{7} \\
1
\end{pmatrix}
\]

Here, \( s \) and \( t \) are free parameters that can take any real value.

Would you like any further details or clarifications?

Here are some related questions you might find helpful:
1. How can you verify the solution of a system of linear equations?
2. What are the geometric interpretations of solutions to linear systems?
3. How does the rank of a matrix relate to the number of solutions?
4. What methods are there for solving systems of linear equations?
5. What are the implications of having free variables in a solution?

**Tip:** When solving systems of equations, organizing your work clearly can help prevent errors, especially in substitution and simplification steps.

Solve the system of equations: x1 - 5x2 + 2x3 - 3x5 - 3x6 = -1, x1 - 5x2 = 0, -x4 - 3x5 - 2x6 = -3, -7x5 - 5x6 = -3.

This guide provides a step-by-step solution to a system of linear equations involving six variables and two free parameters. Learn how to express the solution in parametric form using Gaussian elimination and back substitution. Suitable for high school students in Grades 10-12.

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