<Question>
If $|x+5|-|3x-6|=0$, find the largest possible value of $x$. Express your answer as an improper fraction.
</Question>

<think>
<Plan_1>
Okay, so I need to solve the equation |x + 5| - |3x - 6| = 0 and find the largest possible value of x. Hmm, absolute value equations can sometimes be tricky because they involve multiple cases. Let me start by understanding what this equation is saying.

<prompt_1.1> Rewrite the given equation in a more manageable form as described in the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The equation is |x + 5| = |3x - 6|. That's because if |A| - |B| = 0, then |A| = |B|. So, the absolute value of (x + 5) is equal to the absolute value of (3x - 6). Absolute value equations like this usually require considering different cases where the expressions inside the absolute values are either positive or negative. 

First, let me recall that |A| = |B| implies that either A = B or A = -B. So, applying that here, we can split this into two separate equations:

1. x + 5 = 3x - 6
2. x + 5 = -(3x - 6)

I need to solve both of these equations for x and then check which solutions satisfy the original equation. But before I proceed, I should also consider the critical points where the expressions inside the absolute values change their sign. That is, when x + 5 = 0 and when 3x - 6 = 0. These points are x = -5 and x = 2, respectively. These critical points divide the real number line into different intervals where the expressions inside the absolute values maintain consistent signs. Therefore, I should analyze each interval separately to determine the correct form of the absolute value expressions.

The critical points are x = -5 and x = 2. So, the intervals are:

1. x < -5
2. -5 ≤ x < 2
3. x ≥ 2

In each interval, I can remove the absolute value signs by considering the sign of each expression inside them.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me start with the first interval: x < -5.

In this interval, x is less than -5. Let's check the expressions inside the absolute values:

<prompt_2.1> Solve the equation for the interval x < -5 as described in the above plan. </prompt_2.1>

Moving on to the second interval: -5 ≤ x < 2. Here, x is between -5 and 2.

<prompt_2.2> Solve the equation for the interval -5 ≤ x < 2 as described in the above plan. </prompt_2.2>

Now, the third interval: x ≥ 2. Here, x is 2 or greater.

<prompt_2.3> Solve the equation for the interval x ≥ 2 as described in the above plan. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
1. x + 5: Since x < -5, x + 5 < 0. So, |x + 5| = -(x + 5) = -x -5.
2. 3x - 6: Since x < -5, 3x < -15, so 3x -6 < -21. Therefore, 3x -6 is negative. Hence, |3x -6| = -(3x -6) = -3x +6.

So, the equation |x +5| = |3x -6| becomes:

-x -5 = -3x +6

Let's solve this equation:

Add 3x to both sides: 2x -5 = 6

Add 5 to both sides: 2x = 11

Divide by 2: x = 11/2 = 5.5

But wait, we need to check if this solution is within the current interval, which is x < -5. However, x = 5.5 is not less than -5. So, this solution is extraneous and we must discard it. Therefore, in the interval x < -5, there is no solution.
</execution_2.1>

<execution_2.2>
For x + 5: When x ≥ -5, x + 5 ≥ 0, so |x + 5| = x + 5.

For 3x - 6: When x < 2, 3x - 6 < 0, so |3x - 6| = -(3x - 6) = -3x + 6.

Therefore, in this interval, the equation |x + 5| = |3x - 6| becomes:

x + 5 = -3x + 6

Let me solve this equation.

x + 5 = -3x + 6

Add 3x to both sides:

4x + 5 = 6

Subtract 5 from both sides:

4x = 1

Divide by 4:

x = 1/4

But wait, this interval is -5 ≤ x < 2. The solution x = 1/4 is within this interval, so it's valid.
</execution_2.2>

<execution_2.3>
Let's check the expressions here:

Okay, so now I need to consider each interval, write the equation without absolute values accordingly, solve for x, and check if the solution lies within the interval. Then, among all valid solutions, pick the largest one.

Let me proceed step by step.

**Case 1: x < -5**

In this interval, x is less than -5. Let's check the expressions inside the absolute values:

1. x + 5: Since x < -5, x + 5 < 0.
2. 3x - 6: Since x < -5, 3x < -15, so 3x - 6 < -21, which is negative.

Therefore, both expressions inside the absolute values are negative here. So, |x + 5| = -(x + 5) and |3x - 6| = -(3x - 6). Therefore, the equation |x + 5| = |3x - 6| becomes:

-(x + 5) = -(3x - 6)

Let me simplify this:

Left side: -x -5

Right side: -3x +6

So:

-prompt_2.1.1> Solve the equation for x in the interval x < -5 and check if the solution is valid as described in the above plan. </prompt_2.1.1>
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me check if x = 1/4 is indeed in this interval.

Yes, 1/4 is 0.25, which is between -5 and 2. So, x = 1/4 is a valid solution.

Now, moving to the third case.
-----

So, the solutions are x = 1/4 and x = 5.5. Wait, but earlier when I considered x >=2, I found x=5.5. But I need to check if this is valid in that interval.

Since x=5.5 is indeed greater than or equal to 2, it's within the interval x >=2. So, x=5.5 is a valid solution.

Therefore, the two solutions are x=1/4 and x=11/2. The problem asks for the largest possible value of x.

Comparing 1/4 (which is 0.25) and 11/2 (which is 5.5), the largest value is 11/2.
Therefore, the answer should be 11/2.

But let me just verify both solutions in the original equation to make sure there were no mistakes in the algebra.

<prompt_3.1> Verify both solutions in the original equation as described in the above plan. </prompt_3.1>
-----

So, the answer is 11/2.

But wait, let me just check if there are any other possible solutions or if I missed any cases.

<prompt_3.2> Confirm the solution and ensure no other solutions were missed as described in the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
First, x = 1/4:

|1/4 + 5| - |3*(1/4) -6| = |1/4 + 20/4| - |3/4 - 24/4| = |21/4| - |-21/4| = 21/4 - 21/4 = 0. Correct.

Now, x = 11/2 = 5.5:

|5.5 +5| - |3*5.5 -6| = |10.5| - |16.5 -6| = 10.5 - 10.5 = 0. Correct.

So both solutions are valid. Therefore, the largest possible value of x is 11/2.
</execution_3.1>

<execution_3.2>
The critical points are x = -5 and x = 2, which divide the number line into three intervals. We considered all three cases. Each case was solved, and only the solutions within the intervals were accepted. Therefore, the solutions are x=1/4 and x=11/2. Since the problem asks for the largest possible value of x, 11/2 is the answer.
</execution_3.2>
</Execution_3>

<Final_answer>
Among these values, the largest possible value of $x$ is $\boxed{\frac{11}{2}}$.
</Final_answer>
</think>