Problem: If for any three distinct numbers a,ba, ba,b and ccc we define a,b,c\boxed{a, b, c}a,b,c by
a,b,c=c+ac−b,\boxed{a, b, c}=\dfrac{c+a}{c-b}, a,b,c=c−bc+a,
then 1,−2,−3=\boxed{1,-2,-3}=1,−2,−3=
Answer Choices:
A. −2-2−2
B. −25-\dfrac{2}{5}−52
C. −14-\dfrac{1}{4}−41
D. 25\dfrac{2}{5}52
E. 222
Solution:
If a=1,b=−2a=1, b=-2a=1,b=−2 and c=−3c=-3c=−3, then a,b,c=c+ac−b=−3+1−3+2=−2−1=2\boxed{a, b, c}=\dfrac{c+a}{c-b}=\dfrac{-3+1}{-3+2}=\dfrac{-2}{-1}=2a,b,c=c−bc+a=−3+2−3+1=−1−2=2.