Selesaikan setiap ketaksamaan linear serentak yang berikut. Solve each of the following simultaneous linear inequalities.

$$2x - 7 + 7 > 9 + 7$$ $$2x > 16$$

$$\frac{2x}{2} > \frac{16}{2}$$ $$x > 8$$

Second inequality: $6 - x < 8$

$$6 - x - 6 < 8 - 6$$ $$-x < 2$$

$$-x \times (-1) > 2 \times (-1)$$ $$x > -2$$

Combining the solutions $x > 8$ and $x > -2$, the common region is where both conditions are met. If $x$ must be greater than 8, it is automatically greater than -2.
The solution is $\boxed{x > 8}$.

2. $4x - 5 > 15$ and $3x + 8 > 44$

First inequality: $4x - 5 > 15$

$$4x - 5 + 5 > 15 + 5$$ $$4x > 20$$

$$\frac{4x}{4} > \frac{20}{4}$$ $$x > 5$$

Second inequality: $3x + 8 > 44$

$$3x + 8 - 8 > 44 - 8$$ $$3x > 36$$

$$\frac{3x}{3} > \frac{36}{3}$$ $$x > 12$$

Combining the solutions $x > 5$ and $x > 12$, the common region is where both conditions are met. If $x$ must be greater than 12, it is automatically greater than 5.
The solution is $\boxed{x > 12}$.

3. $1 - 4x \ge 17$ and $2x > 5x + 6$

First inequality: $1 - 4x \ge 17$

$$1 - 4x - 1 \ge 17 - 1$$ $$-4x \ge 16$$

$$\frac{-4x}{-4} \le \frac{16}{-4}$$ $$x \le -4$$

Second inequality: $2x > 5x + 6$

$$2x - 5x > 5x + 6 - 5x$$ $$-3x > 6$$

$$\frac{-3x}{-3} < \frac{6}{-3}$$ $$x < -2$$

Combining the solutions $x \le -4$ and $x < -2$, the common region is where both conditions are met. If $x$ must be less than or equal to -4, it is automatically less than -2.
The solution is $\boxed{x \le -4}$.

4. $6x + 4 \ge 2x - 8$ and $6 - 2x \ge 4 - x$

First inequality: $6x + 4 \ge 2x - 8$

$$6x + 4 - 2x \ge 2x - 8 - 2x$$ $$4x + 4 \ge -8$$

$$4x + 4 - 4 \ge -8 - 4$$ $$4x \ge -12$$

$$\frac{4x}{4} \ge \frac{-12}{4}$$ $$x \ge -3$$

Second inequality: $6 - 2x \ge 4 - x$

$$6 - 2x + x \ge 4 - x + x$$ $$6 - x \ge 4$$

$$6 - x - 6 \ge 4 - 6$$ $$-x \ge -2$$

$$-x \times (-1) \le -2 \times (-1)$$ $$x \le 2$$

Combining the solutions $x \ge -3$ and $x \le 2$, the common region is where $x$ is greater than or equal to -3 AND less than or equal to 2.
The solution is $\boxed{-3 \le x \le 2}$.