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  1. #1
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    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
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  2. #2
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    Quote Originally Posted by karol.rolski View Post
    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.
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  3. #3
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    Thanks for quick answer but where x-3 = 0 ? I can't see it .


    K
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  4. #4
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    Quote Originally Posted by karol.rolski View Post
    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".
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  5. #5
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    Quote Originally Posted by karol.rolski View Post
    ...
    -(x-3)=x(x-3) / (x-3)
    ...
    Quote Originally Posted by karol.rolski View Post
    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.

    I'll start with the second to last line of your original post:

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

    A product equals zero if one factor equals zero. Thus you get:

    x-3=0~\vee~ x+1=0~\implies~x = 3~\vee~x=-1
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