If a + 1/b = 1 and b + 1/c = 1 for real numbers a, b and c, then what is the value of c + 1/a?

Introduction / Context:
This is an algebraic problem involving three variables connected by two simple rational equations. You are asked to find the value of c + 1/a in terms of these relationships. The structure is designed so that although there are three unknowns and only two equations, the particular combination c + 1/a turns out to be uniquely determined.

Given Data / Assumptions:

• a + 1/b = 1.
• b + 1/c = 1.
• a, b, c are nonzero real numbers.
• Find the value of c + 1/a.

Concept / Approach:
We solve the two given equations successively. First express a in terms of b, then express b in terms of c, and finally compute c + 1/a. The aim is to show that all dependence on c cancels out in this expression, leaving a simple constant value.

Step-by-Step Solution:
Start with a + 1/b = 1. Rearrange: a = 1 − 1/b. Then b + 1/c = 1 ⇒ b = 1 − 1/c. Express a directly in terms of c. Substitute b = 1 − 1/c into a = 1 − 1/b. So a = 1 − 1/(1 − 1/c). Simplify the denominator: 1 − 1/c = (c − 1)/c. Thus 1/(1 − 1/c) = c/(c − 1). Therefore a = 1 − c/(c − 1). Write 1 as (c − 1)/(c − 1) to combine: a = (c − 1)/(c − 1) − c/(c − 1) = −1/(c − 1). So a = −1/(c − 1). Then 1/a = (c − 1)/(−1) = 1 − c. Now compute c + 1/a: c + (1 − c) = 1.
Verification / Alternative check:
Choose any convenient value of c that keeps denominators nonzero, for example c = 2. Then b = 1 − 1/2 = 1/2, and a = 1 − 1/b = 1 − 2 = −1. Now c + 1/a = 2 + 1/(−1) = 1, confirming the derived value.

Why Other Options Are Wrong:
Other options such as 0, 2 or 3 would require different relationships among a, b and c. If you pick sample values consistent with the given equations and compute c + 1/a, you will always obtain 1, never these other numbers. The value −1 would arise from sign mistakes when inverting or combining fractions.

Common Pitfalls:
A common error is to assume that three unknowns and two equations mean that nothing can be determined, but here a particular combination is uniquely fixed. Another mistake is sloppy fraction manipulation when expressing a in terms of c. Writing each algebraic step clearly avoids such issues.

Final Answer:
The required value of c + 1/a is 1.