students assign variables like this:
one system of equations is
+ B = 3C + 2D
6B = D + A + C
A + D = 2B + C
It might appear that the problem cannot be solved because there are
4 unknowns and just 3 equations.
Using the variables above, A = ,
C = 2B, D =.
For B = 4, A:B:C:D = 13:4:8:3
3 for the weight of the cone, the left side of the fourth balance equals
12. Since B + C also equals 12, the solution is 2 objects,
a sphere and a cube.