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2005 AMC 10A Problem 2

Problem 2 of 25EasierAlgebra

For each pair of real numbers a≠b,a \neq b, define the operation ⋆\star as (a⋆b)=a+ba−b. (a \star b) = \frac{a+b}{a-b}. What is the value of ((1⋆2)⋆3)?((1 \star 2) \star 3)?

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Solution

First (1⋆2)=1+21−2=3−1=−3.(1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3. Then (−3⋆3)=−3+3−3−3=0−6=0.(-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0. Thus, the correct answer is C.

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Concepts: custom operation · fraction · order of operations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.