Finding the error in algebraic steps leading to 1=2

[waqarrashid33]

if we get:

a+2b=2a+b
=>a-b=2a-2b
=>a-b=2(a-b)
=>1=2

I can't figure out the problem in it..please tell me

 

[micromass]

Does 2*0=3*0 also imply 2=3??

 

[HallsofIvy]

if we get:

a+2b=2a+b

So 2b- b= 2a- a which gives a= b.

=>a-b=2a-2b
=>a-b=2(a-b)
=>1=2

Since a= b, a- b= 0 and you cannot divide by 0.0

I can't figure out the problem in it..please tell me

 

[micromass]

Note that the law

$$ax=bx~\Rightarrow~a=b$$

Only holds if x is nonzero!

The correct way in solving this is

$$ax=bx~\Rightarrow~ ax-bx=0~\Rightarrow~(a-b)x=0~\Rightarrow ~a-b=0~\text{or}~x=0$$