 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Symmetric expressions (Posted on 2023-09-13) Let a, b, c, x, y and z be numbers such that a=(b+c)/(x-2), b=(c+a)/(y-2), c=(a+b)/(z-2). If xy+yz+zx=1000 and x+y+z=2016, find the value of xyz.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) my take Comment 2 of 2 | This is essentially a system of 5 equations for 6 unknowns, so we shouldn't expect there to be a unique solution, but possibly xyz is unique.

The last two equations have the three unknowns whose product we seek.  There are many solutions to this including these two (x and y approximate):

{-0.5, -2015.5, 0}
{-2636.3, 620.3, 1000}

These have different values for the product xyz.  So what can be said about which one is correct?  That should be where the three original equations come in.  For given x,y,z they can be written as linear in a,b,c:

(2-x)a+b+c=0
a+(2-y)b+c=0
a+b+(2-z)c=0

For which a=0, b=0, c=0 is always a solution.  So these equations can't tell us if our xyz is correct.

 Posted by Jer on 2023-09-14 10:09:00 Please log in:

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