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jamielz
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(3508)^2 - (3510 x 3508) = ? 14 Mar 2007, 20:25
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Got this problem out of the Arco book. I got it wrong and am trying to understand why.

I originally did the following

= (3508)(3508) - (3510 x 3508)
= (3508)(3508) + -3510 x -3508
(divide by 3508)
= 3508 + - 3510
= -2

However the actual answer is -7016

Where is the flaw in my math? When do you cancel across terms? Am I violating some parenthesis rules?



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(3508)^2 - (3510 x 3508) = ? 14 Mar 2007, 21:45
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you cant divide across terms if they are getting added like that

the correct way to approach this problem is

(3508)(3508) - (3510 x 3508)

you can break up the terms as follows

(3508)(3508) - (3508)(3508) - (2*3508) = -7016
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Amardeep Sharma
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(3508)^2 - (3510 x 3508) = ? 14 Mar 2007, 23:07
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also there is a flaw in your step number two, though you have corrected the same later...

-(a*b) = -ab and not + - a*-b (else it will become ab instead of -ab)

rest is as explained by terp

regards,

Amardeep
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jamielz
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(3508)^2 - (3510 x 3508) = ? 15 Mar 2007, 11:03
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Terp and Amardeep, thank you

I'm trying to find the alebra principle behind why I cannot cancel out 3508. What I see is

(a x a) - (a * a +2)

= a - a+2

Why am I wrong in cancelling out 3508 from boths sides of the plus sign? Im trying to identify why I can't do it (I know that I can't do it).
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amd08
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(3508)^2 - (3510 x 3508) = ? 15 Mar 2007, 12:28
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(a x a) - (a * a +2)

= a - a+2 is not right.

The equation is actually ->

(a * a) - (a * (a+2))
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Summer3
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(3508)^2 - (3510 x 3508) = ? 15 Mar 2007, 18:36
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This is how I solved it:

(3508^2) - (3510*3508)
= (3508*3508) - (3510*3508)

By taking 3508 common
= 3508 (3508 - 3510)
= 3508 (-2)
= -7016 is the answer.

OR
(a*a) - (b*a) = a (a-b)
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jamielz
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(3508)^2 - (3510 x 3508) = ? 15 Mar 2007, 22:13
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Not trying to belabor the issue, but I guess my real question may not be coming across. What I mean to ask is *why* can't I do this

(3508^2) - (3510*<b>3508</b>)
=(3508 * 3508) - (3510 * 3508)

Then divide this whole baby by 3508, cancelling 3508 on each side of the minus sign, resulting in:

= (3508 * cancel) - (3510 * cancel)
= 3508 - 3510
= -2

I know this is wrong! But dont know WHY.
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Summer3
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(3508)^2 - (3510 x 3508) = ? 16 Mar 2007, 00:12
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(3508^2) - (3510*3508)
=(3508 * 3508) - (3510 * 3508)

You can divide AND multiply this whole with 3508. In the first step divide this whole with 3508 but after you get the answer which is -2, you need to multiply it with 3508...

[{(3508 * 3508) - (3510 * 3508)} * 3508] / 3508

= {(3508 * cancel) - (3510 * cancel)} * 3508
= (3508 - 3510) * 3508
= -2 * 3508
= -7016

Hope it helps!
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(3508)^2 - (3510 x 3508) = ? 16 Mar 2007, 00:37
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jamielz
Not trying to belabor the issue, but I guess my real question may not be coming across. What I mean to ask is *why* can't I do this

(3508^2) - (3510*<b>3508</b>)
=(3508 * 3508) - (3510 * 3508)

Then divide this whole baby by 3508, cancelling 3508 on each side of the minus sign, resulting in:

= (3508 * cancel) - (3510 * cancel)
= 3508 - 3510
= -2

I know this is wrong! But dont know WHY.
Show more


The reason: after you divide, it is the right way to solve this type of question. but after you divide, you need to multiply it again.
so your -2*3508 is the result.
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jamielz
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(3508)^2 - (3510 x 3508) = ? 16 Mar 2007, 07:06
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Vivid. Thank you. I can sleep better tonight.
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javed
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(3508)^2 - (3510 x 3508) = ? 16 Mar 2007, 10:29
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jamielz
Got this problem out of the Arco book. I got it wrong and am trying to understand why.

I originally did the following

= (3508)(3508) - (3510 x 3508)
= (3508)(3508) + -3510 x -3508
(divide by 3508)
= 3508 + - 3510
= -2

However the actual answer is -7016

Where is the flaw in my math? When do you cancel across terms? Am I violating some parenthesis rules?
Show more


Well i would say that you are not wrong is just that you forgot to multiply with the number with which you had divided the equation so taking that in to account the final answer will be then -2 * 3508 = -7016



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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Thank you for understanding, and happy exploring!