Problem: Define the operation "β\circβ" by xβy=4xβ3y+xyx \circ y=4 x-3 y+x yxβy=4xβ3y+xy, for all real numbers xxx and yyy. For how many real numbers yyy does 3βy=12?3 \circ y=12?3βy=12?
Answer Choices:
A. 000
B. 111
C. 333
D. 444
E. more than 444
Solution:
Substituting 333 for xxx yields 3βy=12β3y+3y=123 \circ y=12-3 y+3 y=123βy=12β3y+3y=12. Thus 3βy=123 \circ y=123βy=12 is true for all real numbers yyy.