Question 1209890
Let's analyze the given equation:

$$|x + 4| + |x - 7| = 3x - 1 + |3x - 7|$$

We need to consider different cases based on the values of $x$ that make the absolute value expressions zero.

**Case 1: $x \le -4$**

* $|x + 4| = -(x + 4) = -x - 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = -(3x - 7) = -3x + 7$

Substitute into the equation:
$$-x - 4 - x + 7 = 3x - 1 - 3x + 7$$
$$-2x + 3 = 6$$
$$-2x = 3$$
$$x = -\frac{3}{2}$$

Since $-\frac{3}{2} > -4$, this case is not valid.

**Case 2: $-4 < x \le \frac{7}{3}$**

* $|x + 4| = x + 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = -(3x - 7) = -3x + 7$

Substitute into the equation:
$$x + 4 - x + 7 = 3x - 1 - 3x + 7$$
$$11 = 6$$
This is a contradiction, so there are no solutions in this case.

**Case 3: $\frac{7}{3} < x \le 7$**

* $|x + 4| = x + 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = 3x - 7$

Substitute into the equation:
$$x + 4 - x + 7 = 3x - 1 + 3x - 7$$
$$11 = 6x - 8$$
$$19 = 6x$$
$$x = \frac{19}{6}$$
Since $\frac{7}{3} = \frac{14}{6}$ and $\frac{19}{6} > \frac{14}{6}$, and $\frac{19}{6} = 3\frac{1}{6} < 7$, $x = \frac{19}{6}$ is a solution.

**Case 4: $x > 7$**

* $|x + 4| = x + 4$
* $|x - 7| = x - 7$
* $|3x - 7| = 3x - 7$

Substitute into the equation:
$$x + 4 + x - 7 = 3x - 1 + 3x - 7$$
$$2x - 3 = 6x - 8$$
$$5 = 4x$$
$$x = \frac{5}{4}$$
Since $\frac{5}{4} < 7$, this case is not valid.

**Check the solution $x = \frac{19}{6}$:**

* $|\frac{19}{6} + 4| + |\frac{19}{6} - 7| = |\frac{43}{6}| + |\frac{-23}{6}| = \frac{43}{6} + \frac{23}{6} = \frac{66}{6} = 11$
* $3(\frac{19}{6}) - 1 + |3(\frac{19}{6}) - 7| = \frac{19}{2} - 1 + |\frac{19}{2} - 7| = \frac{17}{2} + |\frac{5}{2}| = \frac{17}{2} + \frac{5}{2} = \frac{22}{2} = 11$

Therefore, the solution is $x = \frac{19}{6}$.

Final Answer: The final answer is $\boxed{\frac{19}{6}}$