Where did C go?!?

Question

Take the equation below, trace the steps, and tell me what happened to variable C.

A = B + C (multiply both sides by (A - B))

A(A - B) = (A - B)(B + C) (multiply this out)

A^2 - AB = AB + AC - B^2 - BC (now subtract AC from both sides)

A^2 - AB - AC = AB - B^2 - BC (now simply each side by factoring)

A(A - B - C) = B(A - B - C) (now divide both sides by (A - B - C)

A = B (what happened to C?! My original equation stated that A = B + C, not that A = B. Where did C go?)

Answer 1

If A = B + C, then A - B = C, and (A - B - C) = C - C = 0, and you have ended up dividing by 0 which is not allowed.

Answer 2

You are dividing by zero

(A - B - C) = 0

Since

A = B + C

Hence,

A - B - C = 0

When you divide by zero, you can equate anything.
{Hence the answer is indeterministic and not useful}


A (A - B - C) = B (A - B - C)

A (0) = B (0)

e.g. 123 (0) = 4 (0) is also true.

You can then claim,

123 = 4
{obviously, this is false}

The value of C when A = B, is C = 0.

But this is immaterial since A = B is erroneous in the first place when you divide by zero by cancelling out the term (A - B - C).

Answer 3

You can't divide by 0 in mathematics.

A = B + C,
so A - B - C = 0.

Not allowed. Bad bad bad. Don't do it.

Answer 4

Hi,
in your original equation:
A=B+C or A-B=C
At the end of the statements,
In short, you are saying
A=B
=> A-B=0 or =C=0

you need to think that...

Answer 5

The answer is simple A-B-C=0 .
you mustn't divide by 0.

Answer 6

It's magic...

HAHAHAHA... just kidding...

You can't just divide the (A-B-C) because according to the first equation, A= B+C.

Rearrange this equation, A-B = C, once more... A-B-C = 0

And you can't divide by a zero...

Answer 7

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