4x+y+2z=100, x>y, y>z, x+y+z=j, 3x+z=k, j>k, j+k=100.
July 22, 2014 6:07 PM Subscribe
4x+y+2z=100, x>y, y>z, x+y+z=j, 3x+z=k, j>k, j+k=100. WolframAlpha can only give me alternative forms or a solution involving 'real and imaginary parts of z'. Is there a solution I can count on my fingers?
posted by obiwanwasabi to Science & Nature (13 answers total) 1 user marked this as a favorite
I wish to buy a thing. The thing I buy will have two desirable characteristics: j, and k.
j comprises three sub-characteristics, while k comprises four sub-characteristics.
I have three levels for ranking how important these sub-characteristics are to me - let's say 'super important', 'kinda important' and 'meh, whatever'. Obviously, they are of decreasing importance.
I have rated the three sub-characteristics for j 'super important', 'kinda important' and 'meh, whatever'.
I have rated three of the four sub-characteristics for k 'super important', and the fourth 'meh, whatever'.
This means I have a total of four 'super important', one 'important', and two 'meh, whatevers'.
Finally, j is a little bit more important to me than k. If further defining these variables help, then I think they'd be about a 60/40 split, though I'm flexible. They can be equal if that helps. At this stage I'd settle for k being less than j if it means a solution pops out.
I would like to assign positive numbers to x, y and z, because I'm a bureaucrat, that's why.
j = x + y + z
k = 3x + z
4x + y + z = 100 (per cent)
j + k = 100 (per cent)
I can't change the number of characteristics, sub-characteristics, or levels of importance.
I could make k=2x+y+z, and so change the overall equation to 3x+2y+2z, but I'm not smart enough to know if that helps. It didn't help WolframAlpha.