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Is there a way to solve this?

  1. Aug 2, 2010 #1
    If we know that x + y = z, can we find out what xy = ?

    This is what I did. Say z = 5

    x + y = z
    x + y = 5
    y = 5 - x

    xy = ???
    x(5 - x) = ???
    5x - 2x = ??? (Since 2x = x^2)
    3x = ???

    xy = 3x
    x = xy / 3
    1 = y / 3 (divided by x)
    y = 3

    and since

    x + y = 5
    x + 3 = 5
    x = 2


    xy = (2)(3)
    xy = 6.

    Please tell me if this is valid. Thanks
  2. jcsd
  3. Aug 2, 2010 #2
    I am not clear on your layout: is the value of z known.?.

    If so, say z is the constant k . Still, xy cannot be fully-determined with this

    info alone: basically, you have one equation with one unknown (if z is fixed/known)

    or, worse, if you do not know the value of z, then you have two unknowns and

    one equation.

    One concrete way of looking at the impossibility of finding a unique value for

    xy is this: you have a fence of perimeter, say, 5 (where x+y is half the

    perimeter, but just multiply by 2 ) around a farm .

    Can you determine, using this information alone , uniquely, what the area of the

    farm (which is the expression xy/2) is.?. No: you could have x=4, y=1, then xy=4

    or x=2, y=3, then xy=6.


    Specific Comments:

    First, I don't see how you get the result 2x=x^2; it is not an identity; it holds

    only for x=0 or x=2 .

    Then everything that builds on this cannot work for values x not 0 or 2.

    Check, e.g., the case x=4. Then 2x=8 , and x^2 =16
  4. Aug 2, 2010 #3
    Not valid in general. What you did is just fine if x=2 and z=5, but you would need to know that was true and provide these as constraints along with your original equation.

    Think about it. If you start with 3 unknowns and one equation, you need two more constraints. You decided to insert the constraints of z=5 and 2x=x^2, but it's not clear why you did that, and it's not true in general unless you say so. But, you did say so, so what can we say. By the way, with those constraints, you also have the solution z=5, y=5 and x=0.

    Now if you had an equation like x+1/y=z/y, then you could say that xy+1=z and hence xy=z-1, then for any given value of z, you would know the value of xy.
  5. Aug 2, 2010 #4
    Wow, I need to be more careful. Sorry. Yes we know what the value of Z is.
  6. Aug 3, 2010 #5


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    Science Advisor
    Homework Helper

    No. One possibility is xy = z - 1 and another possibility is xy = 2z-4. (If you insist, a third is xy = 3z-9; at least two of these will be distinct.)
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