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# 4=3... Where did I go wrong?

## Question

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)

## Recommended Posts

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if a +b = c, then a +b -c = 0;

you're dividig by zero

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Lol. I think I've seen a variation on this one. . . and the consequences such as this:

http://media.photobu...e-by-zero-1.jpg
Edited by Thalia
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lol, that was really good, i'm going to use that

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lol, that was really good, i'm going to use that

to fool the non-mathematical variety of people XD

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another version:

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

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Good one

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I like these ones. Sometimes it's better when the equations are more complicated because it hides the division by zero. Good puzzle!

Edited by Morningstar
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Run Time exception : DIVIDE BY ZERO

a+b=c => a+b-c = 0

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good puzzles

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