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

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

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Updated on: 11 Feb 2013, 15:22
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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

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Originally posted by emmak on 11 Feb 2013, 10:54.
Last edited by Bunuel on 11 Feb 2013, 15:22, edited 1 time in total.
Edited the question.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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11 Feb 2013, 15:22
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13
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.).

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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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12 Mar 2013, 05:08
1
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.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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11 Jun 2013, 20:47
Bunuel wrote:
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 $$a=c$$.

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.

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?
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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11 Jun 2013, 21:21
2
dhlee922 wrote:
Bunuel wrote:

$$ab = bc$$ --> $$ab-bc=0$$ --> $$b(a-c)=0$$--> $$b=0$$ or $$a=c$$.

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.

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

The point here is that analyzing the initial equation ab = bc, we know that either b = 0 or a = c.
Plugging in either of these values results in b(a-c)(a+c) being equal to zero.
I think you're confusion comes from extracting "a = -c" from "b(a-c)(a+c) = 0" as opposed to extracting information from the original equation ab = bc (which is what we need to do).

dhlee922 wrote:
Also, why do all of the possible answers equaling zero in choice B mean that B must be true?

Continuing from above:
So recall that from the first equation, b = 0 or a = c.
The expanded term b(a-c)(a+c) is an alternate expression for the second option; i.e, the fact that both of the possible necessary truths (b = 0; a = c) lead to b(a-c)(a+c) = 0 means that the equivalent expression, a^2 * b = b * c^2 must also be true.

I hope that's clear!
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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12 Jun 2013, 15:29
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?
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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12 Jun 2013, 15:34
dhlee922 wrote:
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.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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12 Jun 2013, 15:38
dhlee922 wrote:
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, 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.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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12 Jun 2013, 15:49
yes, thank you everyone for your replies! geez, i find this one to be pretty tricky, but it's only 600-700 level, ugh
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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11 Oct 2018, 12:11
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!
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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20 Aug 2019, 15:01
Official Explanation:

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.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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31 Aug 2019, 09:27
mattce wrote:
dhlee922 wrote:
Bunuel wrote:

$$ab = bc$$ --> $$ab-bc=0$$ --> $$b(a-c)=0$$--> $$b=0$$ or $$a=c$$.

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.

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

The point here is that analyzing the initial equation ab = bc, we know that either b = 0 or a = c.
Plugging in either of these values results in b(a-c)(a+c) being equal to zero.
I think you're confusion comes from extracting "a = -c" from "b(a-c)(a+c) = 0" as opposed to extracting information from the original equation ab = bc (which is what we need to do).

dhlee922 wrote:
Also, why do all of the possible answers equaling zero in choice B mean that B must be true?

Continuing from above:
So recall that from the first equation, b = 0 or a = c.
The expanded term b(a-c)(a+c) is an alternate expression for the second option; i.e, the fact that both of the possible necessary truths (b = 0; a = c) lead to b(a-c)(a+c) = 0 means that the equivalent expression, a^2 * b = b * c^2 must also be true.

I hope that's clear!

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.
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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29 Sep 2019, 13:04
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

Bunuel wrote:
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.).

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Joined: 16 Jun 2019
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Re: For integers a, b, and c, if ab = bc, then which of the foll  [#permalink]

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29 Sep 2019, 13:06
I have the same question

dhlee922 wrote:
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?
Re: For integers a, b, and c, if ab = bc, then which of the foll   [#permalink] 29 Sep 2019, 13:06
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# For integers a, b, and c, if ab = bc, then which of the foll

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