An alternative approach would to recognize that the one's digit of A + B must equal to 3. Since we know for SURE B can only be 5 or 0 (multiple of 5), then A must have a units digit of (5 or 8). Basically if B ended with 0, we would need A to end with 3. Like wise, if B ended with 5 then we would need A to end in 8 to get units digit of 3. 0+3=3/ 5+8=3.
So know the the possible solutions for B are either 4 (7*4=28) or 9 (7*9=63). We can't have A as 9 since we need at least one B.
Try .70(4)+ .5x=6.30.
2.80+5x=6.3.
.5x=3.5
x= 7
7(B)+4(A)= 11 (OE).
