Let's solve the inequality $3x + 11 \leq 6x + 9$ step by step.

Step 1: Simplify the inequality.

Subtract $3x$ from both sides to get: $11 \leq 3x + 9$

Step 2: Isolate $x$.

Now subtract 9 from both sides: $11 - 9 \leq 3x$ $2 \leq 3x$

Step 3: Solve for $x$.

Divide both sides by 3: $\frac{2}{3} \leq x$

Final Solution:

In interval notation, the solution is: $x \in \left[ \frac{2}{3}, \infty \right)$

This means $x$ is greater than or equal to $\frac{2}{3}$.

Would you like further details or have any questions?

Here are some related questions:

1. How do you solve inequalities involving fractions?
2. What happens when you multiply or divide an inequality by a negative number?
3. How can you express the solution of an inequality using a number line?
4. What is the difference between strict and non-strict inequalities?
5. How does interval notation differ from set notation?

Tip: When solving inequalities, always remember to reverse the inequality sign if you multiply or divide by a negative number.