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The problem presents a system of homogeneous equations:

\[
\begin{aligned}
1. &\quad x - y + z - w = 0 \\
2. &\quad 3x + 2y - z + 3w = 0
\end{aligned}
\]

### Step 1: Solve the system of equations

We will solve this system using standard techniques for linear algebra. First, express one variable in terms of others and substitute between the two equations to simplify.

**Equation 1:**
\[
x - y + z - w = 0 \quad \Rightarrow \quad x = y - z + w
\]

Substitute this into **Equation 2**:

\[
3(y - z + w) + 2y - z + 3w = 0
\]
Expanding this:
\[
3y - 3z + 3w + 2y - z + 3w = 0 \quad \Rightarrow \quad 5y - 4z + 6w = 0
\]
Thus, we get the equation:
\[
5y = 4z - 6w \quad \Rightarrow \quad y = \frac{4z - 6w}{5}
\]

### Step 2: Parametrize the solution
Let \( z = t \) and \( w = s \) where \( t \) and \( s \) are free parameters.

Then, using the equation for \( y \):
\[
y = \frac{4t - 6s}{5}
\]

Now substitute back into the equation for \( x \):
\[
x = y - z + w = \frac{4t - 6s}{5} - t + s
\]
Simplifying:
\[
x = \frac{4t - 6s}{5} - \frac{5t}{5} + \frac{5s}{5} = \frac{-t - s}{5}
\]

### Step 3: Solution vectors
Thus, the solution can be written as a linear combination of free parameters \( t \) and \( s \):
\[
\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = t \begin{pmatrix} -\frac{1}{5} \\ \frac{4}{5} \\ 1 \\ 0 \end{pmatrix} + s \begin{pmatrix} -\frac{1}{5} \\ -\frac{6}{5} \\ 0 \\ 1 \end{pmatrix}
\]

### Two basic solutions:
1. For \( t = 1 \) and \( s = 0 \):
   \[
   \begin{pmatrix} -\frac{1}{5} \\ \frac{4}{5} \\ 1 \\ 0 \end{pmatrix}
   \]
2. For \( t = 0 \) and \( s = 1 \):
   \[
   \begin{pmatrix} -\frac{1}{5} \\ -\frac{6}{5} \\ 0 \\ 1 \end{pmatrix}
   \]

These are the two basic solutions to the system.

Would you like further details or have any questions?

Here are 5 related questions you could explore:
1. What is the geometric interpretation of the solution to this system of equations?
2. How can you generalize this method for larger systems of linear equations?
3. What is the role of free parameters in solutions to homogeneous systems?
4. How would the solution change if the system were non-homogeneous?
5. What are the conditions for a homogeneous system to have only the trivial solution?

**Tip:** Homogeneous systems always have at least the trivial solution (where all variables are zero), but non-trivial solutions exist when the system has more unknowns than independent equations.

Find two basic solutions to the following system of homogeneous equations: x - y + z - w = 0 and 3x + 2y - z + 3w = 0.

Solve the system of homogeneous equations x - y + z - w = 0 and 3x + 2y - z + 3w = 0. This problem covers key linear algebra concepts, such as substitution and parametric solutions. Ideal for students in college or university-level mathematics courses.

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