For integers a, b, and c, if ab = bc, then which of the foll

b(a-c)=0 => b=0 or a=c

You can see B in an easier way:
Inequalities hold when we raise them to power on both sides. so a^2= c^2. => b.a^2= b.c^2. Thus B follows right away.

I agree with other explainations by Bunuel.

for answer choice B, could you also say that if a = -c it also equals zero?

Also, why do all of the possible answers equaling zero in choice B mean that B must be true?

i'm still a little confused by the answer choices.

is it because the question stem equates to B=0 OR A=C

so for answer choice A, A doesnt have to equal C because the B equaling 0 will just make the whole expression 0

but then for answer choice B, since B=0 OR A=C, since 1 or the 3 groups in parentheses will be 0, it will make the whole expression 0, therefore B must be true?

Yes, exactly. We know that or b=0 or a-c=0 ( or both) .

We can rewrite B as \(b(a-c)(a+c)\), and since at least one of the terms is 0, the whole expression is 0.

Yes, for A, if b=0, then a may or may not equal to c.

As for B, since b=0 or a=c (a-c=0), then b(a-c)(a+c)=0 must be true, since either the first or the second multiple (or both) is 0.

Hope it's clear.

yes, thank you everyone for your replies! geez, i find this one to be pretty tricky, but it's only 600-700 level, ugh

Hi everyone,

This question is messing with everything I know (which is good). Isn't it legal to divide both parts of the equation by the same value? In this case, by b:

ab = bc --> (ab)/b = (bc)/b --> a = c

What am I doing wrong here?

Thanks!

Official Explanation:
Correct Answer: B

Choices A and C are essentially identical, and should be regarded with suspicion. Upon further review, if b were to equal 0, then a need not necessarily equal b. Try numbers: a = 1, b = 0, c = 2, then (1)(0) = (0)(2), but 1 does not equal 2. Because b could equal zero, answer choice D is also incorrect, as the two sides of the inequality would both equal zero, and therefore be equal. Answer choice E is also not necessarily true, as demonstrated by the trial numbers above. Answer choice B is correct, as it is true both if a = c, or if b is equal to zero.

Clarity did not come to me from this thread until I read your explanation, and I'm very grateful that you took the time to post it.

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

I have the same question

@permalink,

Actually this is very good question, it checks your reasoning skills.
Here is solution -

Given : ab = bc , a,b,c are integers,-
let's solve step by step -
ab = bc ---> b(a-c) = 0 ------> either b = 0 or a - c = 0,

hence ,
option (A) a = c must be true ???? lets' check a-c = 0 how we are sure from given information --- because here is no idea about what is value of b ... may be from given information b = 0 and a = 4 , c = 5 , then how a = c must be true ??? hence..here is no clarity...no this option may be true but not must be true. so option(A) is not correct.

before option(B) will check rest options -

Option(C) a/c = 1 must be ?????? ..how we can are sure.. we don't have any idea about value of b..b may be 0 , and maybe a= 4, c= 7....so a/c =1 ..not possible..hence..here are this option is not correct.

Option(D) abc>bc...????? ...if b= 0..then ???...this option is also not correct.

Option(E) --- out of scope.

let's come for option(B)-

here ba^2 =bc^2 -----> b(a-c)(a+c) = 0 ...and form given information either a=c or b=0..hence...if b= 0 or a -c = 0 then satisfy --lhs =rhs...hence this option must be true.

hence option B is correct.

I hope you understand.

If like my solution please give kudos...

If \(ab = bc\), then

\(ab - bc = 0\)

\(b(a-c) = 0\)

\(b = 0\) or \(a = c\)

Lets look at the statements:

A. Doesn't have to be true. b could equal 0.

B. \(a^2*b=b*c^2\)

\(a^2*b - c^2*b = 0\)

\(b(a^2 - c^2) = 0\)

\(b(a+c)(a-c) = 0\\
\)
Must be true.

C. Doesn't have to be true. b could equal 0.

D. \(abc > bc\)

\(abc - bc >0\)

\(bc(a-1) > 0\)

Not necessarily true.

E. Cleary wrong.

Answer is B.

ab=bc factors to b(a-c)=0. This means b = 0 or a-c =0 ---> a=c, or both arguments are true.

A. a=c not neccessarilly true if b=0 elim
B. a^2b=bc^2---> a^2b-bc^2=0--> b(a^2-c^2) = 0---> b(a+c)(a-c) = 0---> (a+c)*b*(a-c)=0 since we get a scale multiple of our original argument, this must also be true.

OA is B

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