## Elementary Algebra

We will call the length of the first wire x and the length of the second wire y. Thus, we find: $56 = x+y$ And: $(.25x)^2 + (.25y)^2 = 100 \\ 1/16 (x^2 + y ^2) = 100 \\ x^2 + y^2 = 1600$ Plugging in y = 56 - x, we find: $x^2 + (56 -x)^2 = 1600 \\ 2x^2 - 112x +1536 = 0 \\ 2(x-32)(x-24)$ Thus, we find that the lengths are 32 and 24 inches.