# Math Help - interesting question

1. ## 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

2. 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.

3. Thanks for quick answer but where x-3 = 0 ? I can't see it .

K

4. 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".

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

$-(x-3)=x(x-3)~\implies~ 0=x(x-3)+(x-3) ~\implies~ 0= (x-3)(x+1)$
$x-3=0~\vee~ x+1=0~\implies~x = 3~\vee~x=-1$