To solve the equation \( 3ax + 3 = x - 1 \), follow these steps:

1. **Move all terms involving \( x \) to one side**:
   Subtract \( x \) from both sides:
   \[
   3ax - x + 3 = -1
   \]
   This simplifies to:
   \[
   (3a - 1)x + 3 = -1
   \]

2. **Isolate the term with \( x \)**:
   Subtract 3 from both sides:
   \[
   (3a - 1)x = -4
   \]

3. **Solve for \( x \)**:
   Divide both sides by \( 3a - 1 \):
   \[
   x = \frac{-4}{3a - 1}
   \]

Thus, the solution for \( x \) is:
\[
x = \frac{-4}{3a - 1}
\]

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What happens when \( a = 1 \) in the equation \( 3ax + 3 = x - 1 \)?
2. How do you determine if an equation has no solution?
3. How would you solve if the equation were \( 3ax + 3 = x + 1 \) instead?
4. What if the equation included a quadratic term, like \( 3ax^2 + 3 = x - 1 \)?
5. How would the solution change if there were more variables, like \( 3ax + by + 3 = x - 1 \)?

**Tip**: Always check for restrictions in variables (like when a denominator could make the solution undefined).

This solution walks you through solving the linear equation 3ax + 3 = x - 1 using algebraic techniques. It covers key concepts such as isolating the variable x and manipulating terms. Suitable for students in Grades 9-11.

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