The total area of the surfaces of two cubes is 312 square inches, the total length of their edges is 120 inches. Find the length of the edges of each.

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.