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We are given the following system of linear equations with parameters \( m \), \( x \), \( y \), and \( z \):

\[
\begin{cases}
x + my + (m+1)z = 1 \\
mx + y + z = m \\
(1 - m)x + (m-1)y + m^2 z = 0
\end{cases}
\]

### Steps to solve the system:

We will solve this system using substitution or elimination, and find the general solution in terms of \( m \).

1. **First Equation:**
   \[
   x + my + (m+1)z = 1
   \]
2. **Second Equation:**
   \[
   mx + y + z = m
   \]
3. **Third Equation:**
   \[
   (1 - m)x + (m - 1)y + m^2 z = 0
   \]

I will proceed by solving this system systematically.

Let me work through this step-by-step.

### Step 1: Solve the second equation for one variable.

From the second equation:
\[
mx + y + z = m
\]
We can solve for \( z \):
\[
z = m - mx - y
\]

### Step 2: Substitute \( z \) into the first and third equations.

Substitute \( z = m - mx - y \) into the first equation:
\[
x + my + (m+1)(m - mx - y) = 1
\]
Expanding this:
\[
x + my + (m+1)(m - mx - y) = 1 \implies x + my + (m+1)m - (m+1)mx - (m+1)y = 1
\]
Simplifying this:
\[
x + my + m^2 + m - (m+1)mx - (m+1)y = 1
\]
Collect terms for \( x \) and \( y \):
\[
x - (m+1)mx + my - (m+1)y + m^2 + m = 1
\]
Simplifying further:
\[
(1 - m(m+1))x + (m - (m+1))y + m^2 + m = 1
\]
\[
(1 - m^2 - m)x - y + m^2 + m = 1
\]
Thus, the equation becomes:
\[
(1 - m^2 - m)x - y + m^2 + m = 1
\]

### Step 3: Substitute \( z \) into the third equation.

Now, substitute \( z = m - mx - y \) into the third equation:
\[
(1 - m)x + (m - 1)y + m^2(m - mx - y) = 0
\]
Expanding this:
\[
(1 - m)x + (m - 1)y + m^2m - m^3x - m^2y = 0
\]
Simplifying:
\[
(1 - m)x + (m - 1)y + m^3 - m^3x - m^2y = 0
\]
Collect terms:
\[
(1 - m - m^3)x + (m - 1 - m^2)y + m^3 = 0
\]
Thus, we have:
\[
(1 - m - m^3)x + (m - 1 - m^2)y + m^3 = 0
\]

### Step 4: Solve the system of two equations in \( x \) and \( y \).

We now have two equations in \( x \) and \( y \):

1. \( (1 - m^2 - m)x - y + m^2 + m = 1 \)
2. \( (1 - m - m^3)x + (m - 1 - m^2)y + m^3 = 0 \)

Let me solve this system of two equations.

The solution to the system for \( x \) and \( y \) in terms of \( m \) is:

\[
x = \frac{m^3 - m + 1}{m^3 - m}, \quad y = \frac{-m^2 - m + 1}{m^3 - m}
\]

Now that we have the expressions for \( x \) and \( y \), we can substitute these into the expression for \( z \) from earlier:

\[
z = m - mx - y
\]

Let me calculate \( z \) as well.The solution for \( z \) is:

\[
z = \frac{1}{m}
\]

### Summary of the solutions:
\[
x = \frac{m^3 - m + 1}{m^3 - m}, \quad y = \frac{-m^2 - m + 1}{m^3 - m}, \quad z = \frac{1}{m}
\]

Would you like any further details or explanations?

Here are 5 related questions:
1. How would the solution change if \( m = 0 \)?
2. What happens when \( m = 1 \)? Are there any singularities?
3. How can this system be solved using matrix methods?
4. What are the conditions for the system to have a unique solution?
5. Can this system be generalized for other variables or parameters?

**Tip:** Always check for special values of parameters (like 0 or 1) in parameterized systems to ensure that no division by zero or undefined behavior occurs.

Résoudre le problème: Solve the following system of linear equations: 

 x + my + (m+1)z = 1 
 mx + y + z = m 
 (1 - m)x + (m - 1)y + m^2z = 0

This page provides a detailed step-by-step solution for solving a system of linear equations involving variables x, y, z, and a parameter m. The solution uses substitution and elimination methods, with the final result expressed in terms of m. Suitable for students in Grades 10-12.

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