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find out what went wrong

the parts indented are just explaining what happened

a=b

2a=2b

2a(a-b)=2b(a-b)

2a(a-b)=2bxa-2bxb

(expansion)

2a(a-b)+a=2bxa-2bxb+a

(divide both sides except the "+a" by "a-b")

2a+a=2b-2b+a

2a+a=a

3a+a

3=1

Spoiler for hint:

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Spoiler for hint:

0

Spoiler for hint]

Spoiler for hint:

in step 5, you divide by 0 (a-b). On one side it is canceling out the (a-b) which is 0 but on the other, it is already zero.
Edited by HRAEDIUS

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Guest

Also, the last assumption is wrong:

3a=a (I'm pretty sure your + was a typo, since = is the logical operator for the problem and it wouldn't make sense otherwise) does not mean 3=1. a could equal 0, since 3x0=0.

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a=b

2a=2b

2a(a-b)=2b(a-b)

2a(a-b)=2bxa-2bxb

(expansion)

2a(a-b)+a=2bxa-2bxb+a

(divide both sides except the "+a" by "a-b")

2a+a=2b-2b+a

2a+a=a

3a+a

3=1

Division by zero.

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find out what went wrong

the parts indented are just explaining what happened

a=b

2a=2b

2a(a-b)=2b(a-b)

2a(a-b)=2bxa-2bxb

(expansion)

2a(a-b)+a=2bxa-2bxb+a

(divide both sides except the "+a" by "a-b")

2a+a=2b-2b+a

2a+a=a

3a+a

3=1

In step 4 you expanded the right side but not the left side...

if you were to follow simple algebra rules and expand both sides of the equal sign the math would go a little different.

in step 4 we would have

2axa-2axb=2bxa-2bxb

now we continue with all other steps

2axa-2axb+a=2bxa-2bxb+a

now we divide by the zero even though that is incorrect. We are then left with

2a-2a+a=2b-2b+a

a=a

1=1

statement is true.

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In step 4 you expanded the right side but not the left side...

if you were to follow simple algebra rules and expand both sides of the equal sign the math would go a little different.

in step 4 we would have

2axa-2axb=2bxa-2bxb

now we continue with all other steps

2axa-2axb+a=2bxa-2bxb+a

now we divide by the zero even though that is incorrect. We are then left with

2a-2a+a=2b-2b+a

a=a

1=1

statement is true.

According to the rules of algebra, expanding is legal.

It is always true that a(b-c)=ab-ac. This is always true irrespective of the values of a, b, and c.

Edited by brhan

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