To solve the inequality $3x + 10 \leq 6x + 6$, let's proceed step-by-step.

1. Move all terms involving $x$ to one side of the inequality:

   $3x + 10 - 6x \leq 6$

   Simplifying, we get:

   $-3x + 10 \leq 6$

2. Move the constant term to the other side of the inequality:

   $-3x \leq 6 - 10$

   Simplifying further, we obtain:

   $-3x \leq -4$

3. Divide by -3, and remember to reverse the inequality sign because we are dividing by a negative number:

   $x \geq \frac{-4}{-3}$

   $x \geq \frac{4}{3}$

4. Expressing the solution as an interval:

   Since $x \geq \frac{4}{3}$, the solution interval is:

   $\left[ \frac{4}{3}, \infty \right)$

Therefore, the solution to the inequality $3x + 10 \leq 6x + 6$ is $\left[ \frac{4}{3}, \infty \right)$.

Would you like further details on the solution process?

Here are some related questions for further exploration:

1. How does dividing by a negative number affect the inequality sign?
2. What would happen if the inequality was $3x + 10 \geq 6x + 6$?
3. How can we solve similar inequalities with more complex terms?
4. What is the importance of expressing the solution in interval notation?
5. Can you apply these steps to solve $5x - 2 \leq 2x + 1$?

Tip: Always remember to flip the inequality sign when multiplying or dividing by a negative number.