Question 158560
SEVERAL DAYS AGO, I STARTED WORKING ON THIS PROBLEM AND WAS CALLED AWAY.  I AM NOW BACK AND SEE THAT A SOLUTION HAS NOT BEEN PROVIDED. YOU PROBABLY HAVE ALREADY FOUND A SOLUTION BUT I THINK IT'S A GOOD PROBLEM, SO I'LL GIVE YOU MY TWO BITS:

This is yet another example of more unknowns than equations.  Many times, multiple answers are possible.
First, lets deal in ounces:
Let x=amount of almonds @ $6 per lb or $3/8 per ounce
Let y=amount of cashews @ $5 per lb or $5/16 per ounce
Let z=amount of peanuts @ $2 per lb or $1/8 per ounce 

Now we are told that:
x+y+z=16------------------------eq1
and ($ are understood)
(3/8)x+(5/16)y+(1/8)z=4 multiply each term by 16
6x+5y+2z=64-----------------------------------------eq2

Multiply eq1 by 6 and subtract eq2 from it and we get:

y+4z=32  solve for y
y=32-4z-------------------------------eq1a

Now we know a couple of things about this problem:
(1) We cannot have negative values for x, y or z
(2) We cannot have non-zero values for x, y, or z (I'm assuming that there has to be some of each in every packet)

Given the above, we see that in eq1a, 32-4z has to be greater than zero, other wise we will have either negative or a zero value for y, so:
32-4z>0  subtract 32 from each side:
-4z>-32 divide each side by -4 (note: the inequality sign changes)
z<8  in order for y to be positive and non-zero
Now lets plug y=32-4z from eq1a into eq1
x+32-4z+z=16 or
x+32-3z=16 subtract 32 from and add 3z to each side
x=3z-16  now in order for x to be positive and non-zero, 3z-16 must be greater than zero:

3z-16>0  add 16 to each side
3z>16  divide each side by 3
z>5 1/3 in order for x to be positive and non-zero

WE HAVE NOW PLACED CONSTRAINTS ON Z:
(5 1/3)< z <(8). In order for x and y to be positive and non-zero
Now we can start building the table:
Choose ANY value (INFINITE NUMBER DEPENDING ON THE ACCURACY OF THE SCALES) for z within the above constraints, plug the value into eq1a and solve for y, then plug both values into eq1 and solve for x.  Lets try an example:

Let z be 7, then from eq1a, y=4, then from eq1 x+4+7=16
x+16-11; x=5

CK
5+4+7=16
6*5+5*4+7*2=64
30+20+14=64
64=64
Niche problem!!!

Hope this helps---ptaylor