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Old 01-04-2008, 10:48 AM   #1
joepole
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Actually, step 3 is fine, the distributive property was used correctly.

The problem is that between 3 and 4 you divide both sides by (a + b - c) which, according to the initial equation, is 0. You can't divide by 0.
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Old 01-04-2008, 12:54 PM   #2
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Quote:
Originally Posted by joepole View Post
Actually, step 3 is fine, the distributive property was used correctly.

The problem is that between 3 and 4 you divide both sides by (a + b - c) which, according to the initial equation, is 0. You can't divide by 0.
Yes, the distributive property was used correctly, but it shouldn't have been used at all. Combining like terms is a fundamental of algebra and the author avoided it entirely in order to get the outcome that he did.

Go to http://www.algebrahelp.com/calculators/equation/ and run the equation from step 2 and you'll see what I mean.

Improper calculation = invalid outcome!
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Old 01-04-2008, 02:52 PM   #3
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yeah, I like this one though. As JP says...
if a+b=c then (a+b-c) = 0

So in the last step you are actually saying x divided by zero equals y divided by zero:
3/(a+b-c)=4/(a+b-c)
or
3/0=4/0

heh, tricky
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Old 01-04-2008, 04:16 PM   #4
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Quote:
Originally Posted by j.nc View Post
yeah, I like this one though. As JP says...
if a+b=c then (a+b-c) = 0

So in the last step you are actually saying x divided by zero equals y divided by zero:
3/(a+b-c)=4/(a+b-c)
or
3/0=4/0

heh, tricky
Yes, that's another couple of rules of algebra broken.

(a+b-c) would equal zero. That, I can agree with, but if you are to add a 3 to one side of the equation, you must add a 3 to the other in order to keep the equation balanced, getting (3/0=3/0). This, in fact, would be a mathematical impossibility because, as Joe said, you can't divide by zero. However, discussing division is a side issue anyway, because no division took place in this equation nor should it have.

So let's see where the author started. He started with this, (a+b=c), which equals zero when you rearrange it as (a+b-c=0). Now, if you wanted to distribute 3 and 4 across (a+b-c), you can NOT put 3 on one side and 4 on the other as the author did. This imbalances the equation so no wonder he got such an impossible outcome.

When you add an element to an equation, you have to add equally to BOTH sides of the equation in order to keep it balanced, getting this (3*4(a+b-c)=3*4(0)). If (a+b-c) equals zero, then while following proper order of operations, this would work down to (0=0), a perfectly acceptable outcome.
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