We are asked to solve the equation $|x-2| + |x-3| + 2|x-4| = 9$.

1. Problem Description

We are asked to solve the equation $|x-2| + |x-3| + 2|x-4| = 9$.

2. Solution Steps

We consider different cases based on the values of $x$:

Case 1: $x \ge 4$. In this case, $x-2 > 0$, $x-3 > 0$, and $x-4 \ge 0$, so $|x-2| = x-2$, $|x-3| = x-3$, and $|x-4| = x-4$.

Thus, the equation becomes

$x-2 + x-3 + 2(x-4) = 9$

$x-2 + x-3 + 2x - 8 = 9$

$4x - 13 = 9$

$4x = 22$

$x = \frac{22}{4} = \frac{11}{2} = 5.5$

Since $5.5 \ge 4$, $x = 5.5$ is a valid solution.

Case 2: $3 \le x < 4$. In this case, $x-2 > 0$, $x-3 \ge 0$, and $x-4 < 0$, so $|x-2| = x-2$, $|x-3| = x-3$, and $|x-4| = -(x-4) = 4-x$.

Thus, the equation becomes

$x-2 + x-3 + 2(4-x) = 9$

$x-2 + x-3 + 8 - 2x = 9$

$-5 + 8 = 9$

$3 = 9$

This is a contradiction, so there are no solutions in this case.

Case 3: $2 \le x < 3$. In this case, $x-2 \ge 0$, $x-3 < 0$, and $x-4 < 0$, so $|x-2| = x-2$, $|x-3| = -(x-3) = 3-x$, and $|x-4| = -(x-4) = 4-x$.

Thus, the equation becomes

$x-2 + 3-x + 2(4-x) = 9$

$x-2 + 3-x + 8 - 2x = 9$

$9 - 2x = 9$

$-2x = 0$

$x = 0$

However, $2 \le x < 3$, so $x=0$ is not a valid solution.

Case 4: $x < 2$. In this case, $x-2 < 0$, $x-3 < 0$, and $x-4 < 0$, so $|x-2| = -(x-2) = 2-x$, $|x-3| = -(x-3) = 3-x$, and $|x-4| = -(x-4) = 4-x$.

Thus, the equation becomes

$2-x + 3-x + 2(4-x) = 9$

$2-x + 3-x + 8 - 2x = 9$

$13 - 4x = 9$

$-4x = -4$

$x = 1$

Since $x < 2$, $x = 1$ is a valid solution.

Therefore, the solutions are $x = 1$ and $x = 5.5$.

3. Final Answer

The solutions are $x = 1$ and $x = 5.5$.