Hi Bunuel- how come we can't start out by dividing both sides by b? I dont think to set everything equal to 0. Thanks
emmak wrote:
For integers a, b, and c, if ab = bc, then which of the following must also be true?
A. a = c
B. a^2*b=b*c^2
C. a/c = 1
D. abc > bc
E. a + b + c = 0
\(ab = bc\) --> \(ab-bc=0\) --> \(b(a-c)=0\)--> \(b=0\) or \(a=c\).
A. a = c. If \(b=0\), then this option is not
necessarily true.
B. a^2*b=b*c^2 --> \(b(a^2-c^2)=0\) --> \(b(a-c)(a+c)=0\). Now, since \(b=0\) or \(a=c\), then \(b(a-c)(a+c)\) does equal to zero. So, we have that this options must be true.
C. a/c = 1. If \(b=0\), then this option is not
necessarily true.
D. abc > bc. If \(b=0\), then this option is not true.
E. a + b + c = 0. If \(b=0\), then this option is not
necessarily true (if b=0 then a+c can take any value this option is not necessarily true.).
Answer: B.