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Find the length of the edges of each.

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Let x and y be the edge lengths of the cubes. Then, the surface areas of the cubes are 6*x^2 and 6*y^2 (a cube has 6 faces). Since a cube has 12 edges, the edge length sums for each cube are 12*x and 12*y.
 
The problem conditions give (after canceling the units):
6*x^2 + 6*y^2 = 312
12*x + 12*y = 120
 
x^2+y^2 = 52
x + y = 10
 
The usual way of solving this type of system is substitution:
 
y = 10 - x
x^2 + (10 - x)^2 = 52
2*x^2 - 20*x + 100 = 52
2*x^2 - 20*x + 48 = 0
x^2 - 10*x + 24 = 0
(x - 6)(x - 4) = 0
x = 6 or x = 4
Using y = 10 - x gives y = 4 or y = 6.
 
Thus, the cubes have edge lengths 4 and 6. (This is true regardless of which of x and y is assigned the value 4).
 
The 4 cube has surface area 6*4*4 = 96, and the 6 cube has surface area 6*6*6 = 216, for a total of 312.
The 4 cube has edge length sum 12*4 = 48, and the 6 cube has edge length sum 12*6 = 72, for a total of of 120. Thus, the answer checks.