4 * (a+b-c) = 3 * (a+b-c)
after above step........
there is an analytical reasoning method....it goes like this.....
since 4 not = 3 but (a+b-c) = (a+b-c) for all a,b &c
hence 4 * (a+b-c) = 3 * (a+b-c) only when (a+b-c) = 0
which means a+b=c
another way......
4 * (a+b-c) = 3 * (a+b-c)
after this.....
=> 4 * (a+b-c) - 3 * (a+b-c)=0
=> (a+b-c) = 0
=> a+b = c
the way you did was wrong because.......here only value that satifies for (a+b-c) is 0 and you are cutting 0 by 0 on both sides........which generates a falasy..........
there is a rule.........you never just cut away common factors from both sides......because either you introduce a falasy or miss one factor.....eg......
solve this:
(x-2)^2 = (x-2)(x-3)
=> (x-2) = (x-3) {after cutting (x-2) from both sides}
=> no solution since x-2 can never be equal to x-3 for any value of x
look back carefully.....you missed one factor which is (x-2).....because x=2 satisfies the original equation.......
2006-10-07 20:29:20
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answer #1
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answered by DeAd MaN 4
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First see if a similar selection is added each and every time to get the subsequent selection. Like in #3, you maintain including 4 to a selection to get the subsequent selection. meaning the expression 4n (the place n = the situation of the term interior the series) must be interior the equation someplace. If a sub n = the nth term interior the series, you understand its equation is a sub n = 4n + something. To get something, put in a million for n and the respond ought to be 3. yet 4n = 4. so which you should upload -a million, or subtract a million. answer: a sub n = 4n - a million you need to use this methodology on most of the different ones. the only one here this is diverse is #4 by way of fact those numbers are expanded by using a similar element, 2. meaning you have 2 to a means of (n + something). on the grounds that 2^0 = a million, you should function -a million or subtract a million from n. answer: a sub n = 2^(n-a million)
2016-11-27 00:19:13
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answer #2
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answered by ? 4
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No 4 is no equal to 3 you see a+b-c equal to 0. dividing by 0 produces something that is undefined. Or another way to look at it is 0 x any number is 0, therefore the two number that are multiply by a don't necessary equal to each other.
2006-10-08 04:42:51
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answer #3
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answered by dart 2
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The last step is invalid.
At the beginning, you assume that a+b=c, so a+b-c=0. But then you divide by a+b-c in the last step, meaning you are dividing by zero, which is not allowed.
The second-to-last step is basically saying 4*0 = 3*0, which is the case because both sides are zero.
2006-10-07 20:18:12
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answer #4
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answered by James L 5
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since a+b=c
a+b-c=0 so when you divide each side of the equation by a+b-c, you are dividing by 0
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
can be rewritten
4*0=3*0 which is true since 0=0, but when you divide both sides by zero, you get an eroneous answer.
2006-10-11 15:55:28
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answer #5
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answered by yupchagee 7
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a+b-c = 0
Therefore
4*0 = 3*0, which is true, but says nothing about the relationship between 4*1 and 3*1.
Put another way, your last step requires division by 0, which is indeterminate. 0/0 does not equal 1
2006-10-07 20:31:04
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answer #6
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answered by Helmut 7
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Have you got it yet, any number mutilplied by 0 must equal 0.
My 7 year old can tell you that 3 is not equal to 4.
Ken
2006-10-07 20:30:07
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answer #7
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answered by Ken B 3
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If: a + b = c.
then (a + b - c) = 0
4 * 0 = 3 * 0
Get it?
2006-10-07 20:22:04
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answer #8
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answered by Calculus 5
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You are wrong in step 2 itself.
If a+b = c, it does not imply the next step. That substitution is where the equation goes awry.
2006-10-07 22:17:51
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answer #9
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answered by cooldude 3
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What's to explain? You divided by zero.
2006-10-07 20:18:51
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answer #10
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answered by arbiter007 6
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