Okay, so I need to solve the equation |x + 5| - |3x - 6| = 0 and find the largest possible value of x. Hmm, let's start by understanding what this equation means. The absolute value expressions can sometimes be tricky because they involve cases where the inside is positive or negative. I remember that to solve equations with absolute values, I need to consider different intervals where the expressions inside the absolute values change their signs. 

First, let me rewrite the equation to make it clearer:

|x + 5| = |3x - 6|

So, the absolute values of two expressions are equal. That means either the expressions themselves are equal or they are negatives of each other. Therefore, I can set up two separate equations:

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

Then, I need to solve each equation for x and check if the solutions are valid in the original equation. But before that, maybe I should identify the critical points where the expressions inside the absolute values change their sign. That is, find the values of x where x + 5 = 0 and 3x - 6 = 0.

For x + 5 = 0, solving gives x = -5.

For 3x - 6 = 0, solving gives 3x = 6 => x = 2.

These critical points divide the real number line into three intervals:

1. x < -5
2. -5 ≤ x < 2
3. x ≥ 2

In each interval, the expressions inside the absolute values will have consistent signs, so I can remove the absolute values accordingly and solve the equation. Let me try this approach.

Starting with the first interval: x < -5

In this interval, x + 5 is negative (since x < -5), so |x + 5| = -(x + 5). Similarly, 3x - 6: since x < -5, 3x is less than -15, so 3x - 6 is less than -21, which is negative. Therefore, |3x - 6| = -(3x - 6).

So the equation becomes:

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

Let me simplify both sides:

Left side: -x - 5

Right side: -3x + 6

So:

-x - 5 = -3x + 6

Adding 3x to both sides:

2x - 5 = 6

Adding 5 to both sides:

2x = 11

x = 11/2

Wait, but we are in the interval x < -5. 11/2 is 5.5, which is greater than -5. So this solution is not in the interval x < -5. Therefore, no solution in this interval.

Moving to the second interval: -5 ≤ x < 2

Here, x + 5 is non-negative (since x ≥ -5), so |x + 5| = x + 5. Now, 3x - 6: since x < 2, 3x < 6, so 3x - 6 < 0. Therefore, |3x - 6| = -(3x - 6).

So the equation becomes:

x + 5 = -(3x - 6)

Simplify the right side:

x + 5 = -3x + 6

Adding 3x to both sides:

4x + 5 = 6

Subtracting 5:

4x = 1

x = 1/4

Now, check if this is in the interval -5 ≤ x < 2. 1/4 is 0.25, which is between -5 and 2. So this is a valid solution.

Third interval: x ≥ 2

In this interval, x + 5 is positive (since x ≥ 2 > -5), so |x + 5| = x + 5. Also, 3x - 6: since x ≥ 2, 3x ≥ 6, so 3x - 6 ≥ 0. Therefore, |3x - 6| = 3x - 6.

So the equation becomes:

x + 5 = 3x - 6

Simplify:

Subtract x from both sides:

5 = 2x - 6

Add 6 to both sides:

11 = 2x

x = 11/2

Which is 5.5, which is greater than 2. So this is valid in the interval x ≥ 2.

So the solutions we have are x = 1/4 and x = 11/2. The problem asks for the largest possible value of x. Comparing 1/4 and 11/2, clearly 11/2 is larger. So the answer should be 11/2.

Wait, but let me verify these solutions by plugging them back into the original equation to make sure I didn't make a mistake.

First, check x = 1/4:

Left side: |1/4 + 5| - |3*(1/4) - 6| = |5.25| - |0.75 - 6| = 5.25 - |-5.25| = 5.25 - 5.25 = 0. Correct.

Now x = 11/2 = 5.5:

Left side: |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 is 11/2.

But just to make sure there are no other cases or solutions I might have missed. For example, when setting |A| = |B|, the solutions can be A = B or A = -B. Did I consider both cases?

Wait, in my initial approach, I considered the intervals and handled the signs accordingly. But perhaps another way is to split the equation into two cases:

Case 1: x + 5 = 3x - 6

Case 2: x + 5 = - (3x - 6)

Then solve each case and check if the solutions are in the correct domain.

Wait, that might be another systematic way. Let me try that.

Case 1: x + 5 = 3x - 6

Solving:

x + 5 = 3x - 6

Subtract x from both sides:

5 = 2x - 6

Add 6:

11 = 2x

x = 11/2

Now check if this solution is valid in the original equation. Since when x = 11/2, both |x +5| and |3x -6| are positive, so this solution is valid.

Case 2: x + 5 = - (3x - 6)

Which is:

x + 5 = -3x + 6

Add 3x to both sides:

4x + 5 = 6

Subtract 5:

4x = 1

x = 1/4

Check validity: For x = 1/4, x +5 is positive, 3x -6 is negative. So |3x -6| = -(3x -6), which is correct. So this solution is valid.

Therefore, the two solutions are 11/2 and 1/4, with 11/2 being the larger one. So that's the answer.

But let me think again: when I split into cases based on the critical points, I found the same solutions, so that's consistent. Therefore, the answer is 11/2.

Wait, another thought: Is there a possibility that when solving the equation |x +5| = |3x -6|, there are more solutions? For example, could there be a solution where both expressions are zero? Let's check if x +5 = 0 and 3x -6 = 0 can be true at the same time. Solving x +5 = 0 gives x = -5, but 3*(-5) -6 = -15 -6 = -21 ≠ 0. So no, they don't coincide. So no overlapping solutions here.

Alternatively, could there be a solution where x +5 = -(3x -6)? Wait, that's the second case I considered. So no, all possible cases are covered.

Therefore, I think 11/2 is the correct answer. The largest possible value of x is 11/2.

**Final Answer**
The largest possible value of \( x \) is \boxed{\dfrac{11}{2}}.
Therefore, the largest possible value of $x$ is $\boxed{\frac{11}{2}}$.
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