Problem: If aβ and bβ are real numbers, the equation 3xβ5+a=bx+1 has a unique solution x [the symbol a ξ =0 means that a is different from zero]:
Answer Choices:
A. for all aβ and bβ
B. if a ξ =2 b
C. if aξ =6
D. if bξ =0
E. if bξ =3
Solution:
3xβ5+a=bx+1,3xβbx=6βa,x=3βb6βaβ if bξ =3