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To solve the inequality $|x - 4| + 2|x + 3| \leq 11$, we'll consider different cases depending on the sign of $x - 4$ and $x + 3$.

 

Case 1: $x - 4 \geq 0$ and $x + 3 \geq 0$

In this case, both absolute values are positive, so we have:

$$(x - 4) + 2(x + 3) \leq 11$$

$$x - 4 + 2x + 6 \leq 11$$

$$3x + 2 \leq 11$$

$$3x \leq 9$$

$$x \leq 3$$

 

Case 2: $x - 4 \geq 0$ and $x + 3 < 0$

In this case, $x - 4 \geq 0$ and $x + 3 < 0$, so $x$ is between -3 and 4.

$$(x - 4) + 2(-x - 3) \leq 11$$

$$x - 4 - 2x - 6 \leq 11$$

$$-x - 10 \leq 11$$

$$-x \leq 21$$

$$x \geq -21$$

 

Case 3: $x - 4 < 0$ and $x + 3 \geq 0$

In this case, $x - 4 < 0$ and $x + 3 \geq 0$, so $x$ is between -3 and 4.

$$-(x - 4) + 2(x + 3) \leq 11$$

$$-x + 4 + 2x + 6 \leq 11$$

$$x + 10 \leq 11$$

$$x \leq 1$$

 

Case 4: $x - 4 < 0$ and $x + 3 < 0$

In this case, both absolute values are negative, so we have:

$$-(x - 4) + 2(-x - 3) \leq 11$$

$$-x + 4 - 2x - 6 \leq 11$$

$$-3x - 2 \leq 11$$

$$-3x \leq 13$$

$$x \geq -\frac{13}{3}$$

 

Now, let's combine the solutions from each case:

 

- Case 1: $x \leq 3$

- Case 2: $x \geq -21$

- Case 3: $x \leq 1$

- Case 4: $x \geq -\frac{13}{3}$

 

The overlapping intervals are $[-21, 3]$ and $\left[ -\frac{13}{3}, 1 \right]$.

 

Therefore, the values of $x$ satisfying the inequality are $\boxed{[-21, 3] \cup \left[ -\frac{13}{3}, 1 \right]}$.