Let h = the hundreds digit Let t = the tens digit Let u = the units or ones dogit
t = u+2
u = h+5
t+h+u = 18Solve the second for h, u-5=h Substitute (u+2) for t, and (u-5) for h, in t+h+u = 18 (u+2) + (u-5) + u = 18 u+2+u-5+u = 18 3u-3 = 18 3u = 21 u = 7 Substitute 7 for u in u-5 = h 7-5 = h 2 = h Substitute 7 for u in t = u+2 t = 7+2 t = 9 Answer: [h][t][u] = 297 Edwin