The first equation gives 30 A + 4B = 200 OR B = 50 - 15/2 A ---- (1)

the third equation gives C-A = 3 OR C = A + 3 ----- (2)

Since the values of A, B and C is one digit and that too different from each other and zero we can conclude that A should be even. Trying the single digit evens that is 2,4,6,8 we get: for A = 2 , B = 35; A = 4, B = 15; A= 6, B = 5 and for A = 8 we get B = -10. So the only possible value for A is 6 and that gives B as 5 and from equation (2) we get C = 9.

Thus we have, A=6, B=5, C=9. and these values satisfy the third equation.