Problem: If a​ and b​ are real numbers, the equation 3x−5+a=bx+1 has a unique solution x [the symbol a î€ =0 means that a is different from zero]:
Answer Choices:
A. for all a​ and b​
B. if a î€ =2 b
C. if aî€ =6
D. if bî€ =0
E. if bî€ =3
Solution:
3x−5+a=bx+1,3x−bx=6−a,x=3−b6−a​ if bî€ =3