call h the hundreds digit, t the tens digit, and o the ones digit. then the number is 100h+10t+o.
the sum of the digits is 9, our h+t+o=9. if we reverse the digits our new number is 100o+10t+h and this number is 495 larger than the first. we are also given that h+t=½o. put it all together to get
100h+10t+o+495=100o+10t+h
99h-99o=-495
h-o=-5
1h+0t-1o=-5
1h+1t+1o=9
1h+1t-.5o=0
now I put this all in an augmented matrix and write it in reduced row echelon form (I'll spare you the gory details, you can use substitution and elimination if you'd rather). I get the original number to be 126
