For integers a, b, and c, if ab = bc

lheiannie07 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

Can some experts show me how to solve this?

OA B

Hi lheiannie07.

Let's take a look at your question.

First, the question is which of the options listed MUST be true. Hence, it has to be true always.

If we choose a=1, b=0 and c=2, then ab=bc=0. Now

(A) a=c is FALSE.
(B) a^2*b=b*c^2=0 is true, but it has to be true always.
(C) a/c = 1 is FALSE, because a/c=1/2.
(D) abc>bc is FALSE, because abc=0=bc.
(E) a+b+c=0 is FALSE, because a+b+c=1+0+2=3.

Hence, the only option that was true is (B). It implies that the correct option is B.

But, if we want to show that it is always true we have to do the following: let be a, b and c general integers, hence $$\text{if}\ ab=bc\ \text{then}\ \ a\cdot\left(ab\right)=a\cdot\left(bc\right)$$ $$\Leftrightarrow\ a^2b=\left(ab\right)\cdot c$$ $$\Leftrightarrow\ a^2b=\left(bc\right)\cdot c$$ $$\Leftrightarrow\ a^2b=bc^2.$$ I hope it helps you.

I'm available if you'd like a follow up.

Regards.

lheiannie07 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.

Implication:
Either b=0 or a-c=0.
Given this constraint, prove that four of the answer choices DON'T have to be true.

Case 1: b=0, a=2, c=1
In this case, A, C, D and E are not true.
Eliminate A, C, D and E.

The correct answer is B.