1. The problem is to solve the quadratic inequality $$6X + 6 \leq 4X + 6$$. 2. First, we write the inequality clearly: $$6X + 6 \leq 4X + 6$$ 3. Subtract $4X$ and $6$ from both sides to isolate terms with $X$: $$6X + 6 - 4X - 6 \leq 4X + 6 - 4X - 6$$ which simplifies to $$6X - 4X + \cancel{6} - \cancel{6} \leq 0$$ 4. Simplify the left side: $$2X \leq 0$$ 5. Divide both sides by 2 (positive number, so inequality direction stays the same): $$\frac{\cancel{2}X}{\cancel{2}} \leq \frac{0}{2}$$ which gives $$X \leq 0$$ 6. The solution to the inequality is all values of $X$ less than or equal to zero. 7. In interval notation, the solution is $$(-\infty, 0]$$. This means any number less than or equal to zero satisfies the inequality.