Another Math problem

Here's another math problem for everyone to enjoy.

If you can't read it - I wrote it out below...

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Originally Posted by StevenSurprenant

Here's another math problem for everyone to enjoy.

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I am unable to see it. Hopefully I can with my iPhone. I think others will also have trouble.

Unless there's a joke I'm missing... Line 3 where they /-A^2 on one side but not the other. At that point both sides are not equal. You plug in the values and you have

1*5 - 1*4 = (5^2 - 2*4*5 + 4^2) / -4^2

1 = (25 - 40 +16) / 16

1 = 1/16

wrong.

Quote:

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Originally Posted by JohnMichael

I am unable to see it. Hopefully I can with my iPhone. I think others will also have trouble.

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Maybe this will be better?

I'll explain the problem later after everyone interested has a chance to figure it out.

I can't see it either. Can you just type it out?

Let
A=4
B=5
C=1

Equation
C = B – A

[ *(B-A) ]
C(B-A) = (B-A)^2
CB – CA = B^2 – 2AB + A^2

[ -A^2 ]
CB – CA – A^2 = B^2 – 2AB + A^2 – A^2
CB – CA – A^2 = B^2 – 2AB

[ +AB ]
AB + CB – CA – A^2 = B^2 – 2AB +AB
AB + CB – CA – A^2 = B^2 – AB

[ -CB ]
AB + CB – CA – A^2 - CB = B^2 – AB - CB
AB – CA – A^2 = B^2 – CB - AB

[ Factor ]
AB – CA – A^2 = B^2 – CB – AB
A(B – C – A) = B(B – C – A)

[ /(B – C – A) ]
(A(B – C – A)) / (B – C – A) = (B(B – C – A)) / (B – C – A)
A = B

4 = 5
2 X 2 =5

Maybe now it can be read?
I did a few more steps to clarify the process.
Have at it!

Checking the solution in reverse...

A(B-C-A)=B(B-C-A)
AB-AC-A^2=B^2-BC-AB
AB-A^2-B^2+AB=AC-BC
AB-A^2-B^2+AB=C(A-B)
C=(2AB-A^2-B^2)/(A-B)
C=(2AB-A^2-B^2)/(A-B) =(A-B)(B-A)/(A-B)
C=B-A

...which is the original equation

Aha! got it.

You can't do the last step which is to divide by (B-A-C) because B-C-A = 0

You can't divide by zero, because it has no meaning. Hence the nonsensical conclusion 4=5

good one!

Quote:

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Originally Posted by noddin0ff

Aha! got it.

You can't do the last step which is to divide by (B-A-C) because B-C-A = 0

You can't divide by zero, because it has no meaning. Hence the nonsensical conclusion 4=5

good one!

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A=4
B=5
C=1

The initial Equation C = B – A works for the supplied constants and also works for A(B – C – A) = B(B – C – A) , but, as you said, not for (A(B – C – A)) / (B – C – A) = (B(B – C – A)) / (B – C – A)

However,

[ /(B – C – A) ]
(A(B – C – A)) / (B – C – A) = (B(B – C – A)) / (B – C – A)
A = B

Is still algebraically correct, but not with the supplied solution set which does work for the initial equation.

In other words: C = B - A is not the same as A(B – C – A) = B(B – C – A).

You're getting close... Now the question is why?

There doesn't seem to be much interest in this post, so I will post the explanation for those that are.

Extraneous and missing solutions - Wikipedia, the free encyclopedia

Scroll down to: "Extraneous solutions: multiplication"

Case closed!

Sorry. I'm interested, I just haven't had time to sit down and focus on this. It's a busy weekend.

Quote:

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Originally Posted by ForeverAutumn

Sorry. I'm interested, I just haven't had time to sit down and focus on this. It's a busy weekend.

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I enjoyed your post, that's why I posted this. I didn't reply to your post because several members solved it and I didn't want to give away the solution. Anyway, I found it fun.

I guess I don't get the nuance. If the initial equation C = B - A is true, then the value of (B - A - C) is necessarily zero for any values that satisfy C = B - A

I guess what Wiki is saying is that the problem was just a step prior to dividing by (B-A-C) and was really when the equation was arranged to multiply both sides by (B-A-C). As soon as you multiply both side by zero all bets are off?

Quote:

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Originally Posted by noddin0ff

I guess I don't get the nuance. If the initial equation C = B - A is true, then the value of (B - A - C) is necessarily zero for any values that satisfy C = B - A

I guess what Wiki is saying is that the problem was just a step prior to dividing by (B-A-C) and was really when the equation was arranged to multiply both sides by (B-A-C). As soon as you multiply both side by zero all bets are off?

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Multiplying both sides by zero is valid, but when you try to factor out (in the last step) (B-C-A) you get an error since you can't divide by zero.

This only occurs with the given values of A=4, B=5, C=1 which is a valid solution for the original equation and the step A(B – C – A) = B(B – C – A), but become invalid (with the initial solution set) when you try to factor out (B-C-A) in the final step.

If you were to graph the solution set for the original equation C=B-A and A(B – C – A) = B(B – C – A), you would find that they have different solution sets so C=B-A is not equivalent to A(B – C – A) = B(B – C – A)

A(B – C – A) = B(B – C – A) solution set is confined to values that only equal zero for (B – C – A), hence this solution set is invalid for the last step to be valid (division by zero)

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