Solution:

To solve the equation [X/2] [2x/3]=x, we need to understand the greatest integer function. The greatest integer function means that we take the largest integer that is less than or equal to the given number. For example, [3.7] = 3 and [-2.5] = -3.

Approach:

We will solve the given equation by using the following approach:

1. Simplify the given equation.
2. Break the equation into different intervals.
3. Find the solution for each interval.
4. Combine the solutions for all intervals.

1. Simplify the given equation:

The given equation is [X/2] [2x/3]=x.

Let's simplify this equation by assuming that X/2 = a, and 2x/3 = b. Then we have:

[a] [b] = 3a/2.

Now, let's consider two cases:

Case 1: When a and b are both positive or zero.

In this case, we have:

[a] [b] = ab.

Thus, the given equation becomes:

ab = 3a/2.

Therefore, we have:

b = 3/2.

This means that:

2x/3 = 3/2.

Thus, x = 9/4.

Case 2: When a and b are both negative.

In this case, we have:

[a] [b] = (a-1)(b-1).

Thus, the given equation becomes:

(a-1)(b-1) = 3a/2.

Therefore, we have:

b = (3a+2)/(2a-1).

Now, let's consider two sub-cases:

Sub-case 1: When b < />

In this sub-case, we have:

(3a+2)/(2a-1) < />

Thus, we have:

a < -2/3="" or="" a="" /> 1/2.

Sub-case 2: When b > 0.

In this sub-case, we have:

(3a+2)/(2a-1) > 0.

Thus, we have:

-2/3 < a="" />< />

2. Break the equation into different intervals:

Based on the above analysis, we can break the given equation into three intervals:

Interval 1: a < />
Interval 2: -2/3 < a="" />< />
Interval 3: a > 1/2.

3. Find the solution for each interval:

Now, let's find the solution for each interval:

Interval 1: a < />

In this interval, we have:

[b] < />

Thus, there is no solution for this interval.

Interval 2: -2/3 < a="" />< />

In this interval, we have:

[b] = 0.

Thus, we have:

(a-1)(b-1) = 3a/2.

This equation can be simplified as:

-5a + 3 = 0.

Thus, we have:

a = 3/5.

This means that:

X/2 = 3/5.

Thus, we have:

X = 6/5.

Interval 3: a > 1/2.

In this interval, we have:

[b]

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