# interesting question

• Jan 27th 2009, 04:13 AM
karol.rolski
interesting question
Hi ,

There is a simple task to calculate x from 1,5+0,5x = x. Simple!
1,5+0,5x=x
1,5=0,5x
x=3,0

However, colleague of mine come up with another answer:
1,5+0,5x=x /*2
3+x=2x /*(x+1)
3+4x+x^2=2x^2+2x /-(x^2)
3+4x=x^2+2x /-(5x)
3-x=x^2-3x
-(x-3)=x(x-3) / (x-3)
-1 = x

I know it is where he complicates things but above logic seems to be correct.
Where is mistake then ?

Thanks
K
• Jan 27th 2009, 04:20 AM
earboth
Quote:

Originally Posted by karol.rolski
Hi ,

There is a simple task to calculate x from 1,5+0,5x = x. Simple!
1,5+0,5x=x
1,5=0,5x
x=3,0

However, colleague of mine come up with another answer:
1,5+0,5x=x /*2
3+x=2x /*(x+1)
3+4x+x^2=2x^2+2x /-(x^2)
3+4x=x^2+2x /-(5x)
3-x=x^2-3x
-(x-3)=x(x-3) / (x-3)
-1 = x

I know it is where he complicates things but above logic seems to be correct.
Where is mistake then ?

Thanks
K

This kind of "logic" is only possible if one factor equals zero.

Since x - 3 = 0 you divide by zero here ... There exists a thread about this subject at MHF. Search for it.
• Jan 27th 2009, 04:30 AM
karol.rolski
Thanks for quick answer but where x-3 = 0 ? I can't see it .

K
• Jan 27th 2009, 05:02 AM
Rincewind
Quote:

Originally Posted by karol.rolski
Thanks for quick answer but where x-3 = 0 ? I can't see it .

Your original equation has the solution x = 3 as you showed.

On the second last step of your friend's working, you are dividing by (x-3)

Substitute in and you will see you are dividing by zero. When you do that anything can happen or be proven "true".
• Jan 27th 2009, 07:04 AM
earboth
Quote:

Originally Posted by karol.rolski
...
-(x-3)=x(x-3) / (x-3)
...

Quote:

Originally Posted by karol.rolski
Thanks for quick answer but where x-3 = 0 ? I can't see it .

If you divide an equation by a term which contains the variable then you remove one solution.