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4=3... Where did I go wrong?
Posted 25 October 2011 - 06:51 PM
Begin with a simple algebraic equation:
a + b = c
Now, replace each letter with 4 of that letter minus 3 of that letter:
4a - 3a + 4b - 3b = 4c - 3c
Rearrange, so that the 4s are on the left and the 3s on the right
4a + 4b - 4c = 3a + 3b - 3c
Simplify by factoring:
4(a + b - c) = 3(a + b - c)
Then divide each side by a+b-c:
4 = 3
Can someone point out the mistake? (an easy one for maths lovers)
Posted 25 October 2011 - 10:36 PM
Edited by Thalia, 25 October 2011 - 10:37 PM.
Posted 26 October 2011 - 10:03 PM
to fool the non-mathematical variety of people XD
lol, that was really good, i'm going to use that
Posted 15 November 2011 - 09:52 AM
1. a = b
2. a2 = ab (multiplying a on both sides)
3. a2 - b2 = ab - b2 (subtracting b2 from both sides)
4. (a+b)(a-b) = b(a-b)
5. a+b = b (divided both sides by (a-b) )
6. 2b = b ( from equation 1 i.e. replacing a=b)
7. 2 = 1
Posted 22 November 2011 - 12:41 AM
Edited by Morningstar, 22 November 2011 - 12:42 AM.
Posted 16 December 2011 - 10:06 AM
a+b=c => a+b-c = 0
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