Thread: Another Math problem

1. Another Math problem

Here's another math problem for everyone to enjoy.

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

2. Originally Posted by StevenSurprenant Here's another math problem for everyone to enjoy.

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

3. 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.  Reply With Quote

4. Originally Posted by JohnMichael I am unable to see it. Hopefully I can with my iPhone. I think others will also have trouble.
Maybe this will be better?

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

5. I can't see it either. Can you just type it out?  Reply With Quote

6. 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!  Reply With Quote

7. 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  Reply With Quote

8. 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!  Reply With Quote

9. 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!
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?  Reply With Quote

10. 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!  Reply With Quote

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

12. 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.
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.  Reply With Quote

13. 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?  Reply With Quote

14. 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?
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)  Reply With Quote

15. Mua Chung want to say hello

Hello there,

I'm Mua Chung. I am coming from Vietnam. Happy to become a member of forums.audioreview.com community. I would like to socialize as well as learn from the experts here.  Reply With Quote