To solve the equation \( y = x e^y + y \), let's simplify and analyze it.

1. **Step 1: Isolate terms involving \( y \):**
   The equation is:
   \[
   y = x e^y + y
   \]
   Subtract \( y \) from both sides:
   \[
   0 = x e^y
   \]

2. **Step 2: Analyze the resulting equation:**
   Since \( x e^y = 0 \), there are two possible cases:
   - **Case 1:** \( x = 0 \)
   - **Case 2:** \( e^y = 0 \) (However, \( e^y \) is never zero for any real \( y \).)

Thus, the only solution occurs when:
\[
x = 0
\]

3. **Conclusion:**
   If \( x = 0 \), then \( y \) can be any value, as the equation \( y = y \) is always satisfied.

### Final Answer:
The solution to the equation \( y = x e^y + y \) is:
\[
x = 0 \quad \text{and} \quad y \text{ can be any real number}.
\]

Would you like further details on this solution or have any questions?

Here are five related questions for deeper exploration:
1. What is the behavior of \( y = x e^y + y \) near \( x = 0 \)?
2. How can this type of equation be interpreted graphically?
3. What methods are available to solve equations involving \( y \) in both linear and exponential forms?
4. Are there specific cases where \( y \) could depend on \( x \) in a different form?
5. How would this equation change if an additional term were added?

**Tip:** When dealing with equations involving both \( y \) and \( e^y \), isolating terms often helps simplify or reveal special solutions.

Explore how to solve the equation y = x e^y + y. The solution reveals that x = 0 and y can be any real number. This problem involves concepts from algebra and exponential equations, suitable for advanced high school students.

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