Solution:
Given a+b+c+d=12
and abcd=27+ab+ac+ad+bc+bd+cd
then a,b,c and d =?
let a=b=c=d=3
then 3+3+3+3=12
LHS => abcd=3*3*3*3=81
RHS=> 27+3*3+3*3+3*3+3*3+3*3+3*3=81
hence a=b=c=d=12
Solve second eq for a, substitute into first eq, and multiply through by bcd to get:
(…) = 12bcd -27.
The left side has to be positive so eqs have a sol for all b,c,d s.t. 12bcd>27
The above is correct, but it doesn’t work, I tried it- it is necessary but not sufficient. A sufficient conditon is given below following an example.
a+b+c+d=12
abcd=27+a(b+c+d)+b(c+d)+cd.
Example:
Assume a=4, and b=2. Then
1) c+d=6
8cd=27+4(8)+cd
2) 7cd=59
1) and 2) give a quadratic equation with solution c=.756 and d=5.244
In general:
Assume a and b given. Then:
3) c+d=12-(a+b)
(ab-1)cd=27+a(b+(12-(a+b))+b[12-(a+b)], or:
4) cd=k
3) and 4) give a quadratic eq for c which has a solution if (c+d)^2>4k (not very nice).
Topsquark: I pissed in your ear and told you it was raining. My sincerest apologies. Feel free to remove the thanks- I appreciate the gesture.