17(3c + 5d) + 6c - 28d при c + d =2

Произведем следующие преобразования:17(3c + 5d) + 6c - 28d = 17(3c + 3d + 2d) + 6c - 28d = 17(3c + 3d) + 17(2d) + 6c - 28d = 51(c + d) + 34d + 6c - 28d = 51(c + d) + 6d + 6c = 57(c + d);Зная, что:c + d = 2;получаем:17(3c + 5d) + 6c - 28d = 57(c + d) = 57 x 2 = 114.