By even showing set solution

$$\frac{-6}{2} > \frac{2x}{2} \implies -3 > x \implies x < -3$$ $$\frac{2x}{2} > \frac{2}{2} \implies x > 1$$

We need $x < -3$ and $x > 1$. There is no value of $x$ that can satisfy both conditions simultaneously.
The solution is no solution.

$$\boxed{\text{No solution}}$$

Number line representation: An open circle at $-3$ with shading to the left, and an open circle at 1 with shading to the right. There is no overlap.
Set solution: $\emptyset$ (the empty set)

$$\boxed{-4 < x < 2}$$

Number line representation: An open circle at $-4$ and an open circle at 2, with the region between them shaded.
Set solution: $(-4, 2)$

$$\boxed{-3 \le x \le 5}$$

Number line representation: A closed circle at $-3$ and a closed circle at 5, with the region between them shaded.
Set solution: $[-3, 5]$

$$\boxed{-4 < x < 8}$$

Number line representation: An open circle at $-4$ and an open circle at 8, with the region between them shaded.
Set solution: $(-4, 8)$

$$-3 - 3 \le 3x+3 - 3 \le 9 - 3$$ $$-6 \le 3x \le 6$$

$$\boxed{-2 \le x \le 2}$$

Number line representation: A closed circle at $-2$ and a closed circle at 2, with the region between them shaded.
Set solution: $[-2, 2]$

$$\boxed{-1 \le x \le 6}$$

Number line representation: A closed circle at $-1$ and a closed circle at 6, with the region between them shaded.
Set solution: $[-1, 6]$

$$2y+3 - 3 < 11 - 3$$ $$2y < 8$$

$$\boxed{y < 4}$$

Number line representation: An open circle at 4, with the region to the left of 4 shaded.
Set solution: $(-\infty, 4)$

$$2x-5 - 2x > 4x+7 - 2x$$ $$-5 > 2x+7$$

$$-5 - 7 > 2x+7 - 7$$ $$-12 > 2x$$

$$\boxed{-6 > x \quad \text{or} \quad x < -6}$$

Number line representation: An open circle at $-6$, with the region to the left of $-6$ shaded.
Set solution: $(-\infty, -6)$