**Problem Analysis:**

We are given the equation x = p-2 and the equation x/p - x 1/p = 1. We need to find the value of p that satisfies these equations.

**Solution:**

Let's solve the equations step by step to find the value of p.

**Step 1: Substitute the value of x in terms of p**

From the first equation x = p-2, we can substitute the value of x in the second equation to eliminate x.

Substituting x = p-2 in x/p - x 1/p = 1, we get:

(p-2)/p - (p-2)1/p = 1

**Step 2: Simplify the equation**

To simplify the equation, let's first simplify the terms on the left side.

For (p-2)/p, we can rewrite it as 1 - 2/p using the concept of division.

Similarly, for (p-2)1/p, we can rewrite it as (p-2) * 1/p using the concept of multiplication.

Substituting these simplified terms in the equation, we get:

(1 - 2/p) - (p-2) * 1/p = 1

Simplifying further, we get:

1 - 2/p - (p-2)/p = 1

**Step 3: Combine like terms**

To combine like terms, let's first simplify the term (p-2)/p.

For (p-2)/p, we can rewrite it as (p/p) - (2/p) using the concept of division.

Substituting this simplified term in the equation, we get:

1 - 2/p - (p/p) + (2/p) = 1

Simplifying further, we get:

1 - 2/p - 1 + 2/p = 1

**Step 4: Simplify the equation**

To simplify the equation, let's combine like terms.

The terms -2/p and 2/p cancel each other out, leaving us with:

1 - 1 = 1

Simplifying further, we get:

0 = 1

**Step 5: Conclusion**

From the equation 0 = 1, we can see that the equation is not true for any value of p.

Therefore, there is no value of p that satisfies the given equations.

**Verification:**

To verify our solution, let's substitute the value of p in the original equations and check if they hold true.

Substituting p = 0 in the equation x = p-2, we get:

x = 0 - 2

x = -2

Substituting p = 0 in the equation x/p - x 1/p = 1, we get:

(-2)/0 - (-2)1/0 = 1

This equation is undefined as division by zero is not defined in mathematics.

Therefore, our solution is verified as there is no value of p that satisfies the original equations.