Let
x = the number of a-cables
y = the number of b-cables
z = the number of c-cables
Cable a uses 3 black, 3 white and 2 red wires.
Cable b uses 1 black, 2 white and 1 red wire.
Cable c uses 2 black, 1 white and 2 red wires.
Because 120 black, 110 white and 100 red wires were used, therefore the system of equations is
3x + y + 2z = 120 (1)
3x + 2y + z = 110 (2)
2x + y + 2z = 100 (3)
Answer:
3x + y + 2z = 120
3x + 2y + z = 110
2x + y + 2z = 100
where
x,y,z = the number of a, b and c-cables respectively.