To solve the differential equation \( \frac{dy}{dx} = -1 - 3y \), we will use the method of **separation of variables**.

### Step-by-step Solution:

1. **Rewrite the equation**:
   \[
   \frac{dy}{dx} = -1 - 3y
   \]
   This can be rearranged as:
   \[
   \frac{dy}{dx} = -3y - 1
   \]

2. **Separate the variables**:
   Move all terms involving \( y \) to one side and all terms involving \( x \) to the other side:
   \[
   \frac{dy}{-3y - 1} = dx
   \]

3. **Integrate both sides**:
   On the left-hand side, we need to integrate with respect to \( y \), and on the right-hand side, we integrate with respect to \( x \):
   \[
   \int \frac{1}{-3y - 1} \, dy = \int 1 \, dx
   \]

   The integral on the left can be done by substitution. Let:
   \[
   u = -3y - 1, \quad \text{so} \quad du = -3 \, dy
   \]

   Thus, the left side becomes:
   \[
   \int \frac{1}{u} \cdot \frac{du}{-3} = -\frac{1}{3} \int \frac{1}{u} \, du = -\frac{1}{3} \ln |u|
   \]

   Substituting back \( u = -3y - 1 \):
   \[
   -\frac{1}{3} \ln | -3y - 1 | = x + C
   \]

4. **Solve for \( y \)**:
   Multiply both sides by \( -3 \):
   \[
   \ln | -3y - 1 | = -3(x + C)
   \]
   
   Simplify:
   \[
   \ln | -3y - 1 | = -3x - 3C
   \]
   Let \( C_1 = -3C \) (since \( C_1 \) is just another constant), so:
   \[
   \ln | -3y - 1 | = -3x + C_1
   \]

5. **Exponentiate both sides**:
   To eliminate the logarithm, exponentiate both sides:
   \[
   | -3y - 1 | = e^{-3x + C_1} = e^{C_1} e^{-3x}
   \]
   Let \( A = e^{C_1} \) (since \( A \) is just a constant), we have:
   \[
   | -3y - 1 | = A e^{-3x}
   \]

6. **Solve for \( y \)**:
   Remove the absolute value:
   \[
   -3y - 1 = \pm A e^{-3x}
   \]
   Solve for \( y \):
   \[
   y = \frac{-1 \pm A e^{-3x}}{3}
   \]

Thus, the general solution to the differential equation is:
\[
y = \frac{-1 \pm A e^{-3x}}{3}
\]
where \( A \) is an arbitrary constant.

---

Would you like more details on any step? Feel free to ask.

Here are some related questions to explore:
1. How can we determine the constant \( A \) using initial conditions?
2. What does the particular solution

This solution demonstrates how to solve the differential equation dy/dx = -1 - 3y using the separation of variables technique. The solution involves integrating both sides, utilizing logarithmic and exponential properties, and finding the general solution. Suitable for Grades 11-12 or college students.

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