SOLUTION:
905-756=149 and 873-392=481.
EXPLANATION
The given equations can be changed to additions and the question marks replaced by letters:
(1)
1A9+75B=CDE
(2)
48V+W9X=YZ3
The letters A, B, C, D, E, V, W, X, Y, Z each represent a unique digit between 0 and 9.
From these sums we can make some more equations relating the letters algebraically:
SET 0: E=B-1, Y=W+5, W>0.
SET 1: D=A+6, C=8, X=3-V, Z=7.
SET 2: D=A-4, C=9, X=3-V, Z=7.
SET 3: D=A-4, C=9, X=13-V, Z=8.
In Sets 1-3, C and Z take specific values—these are the only values allowed.
In Set 0, the most restrictive condition is Y=W+5 giving a set of values for W and Y denoted by (WY)={16 27 38 49}.
In Sets 1 and 2, (VX)=(XV)={03 12}, so V and X have interchangeable values.
In Set 3, (VX)=(XV)=(76) only because 9 and 8 are already assigned to C and Z. This is a constraint for A and D: (AD)={40 51}. So for Set 3 we have the sextets (ACDVXZ)={490768 591768}. However, we can’t build (WY) into either of these. That means we can eliminate Set 3 as part of the solution.
In Sets 1 and 2, we already have (VX). Consider (AD) for Set 1:
(AD)={06 39}; for Set 2 (AD)={06 28}.
The quartet (ADVX)={0612 3912} and the sextet for Set 1 is:
(ACDVXZ)=(086127). This time we can build (WY) into this:
(ACDVWXYZ)=(08614297). The two remaining digits are 3 and 5. These do not fit the requirements for B and E, so we reject Set 1.
We are left with Set 2, in which (AD)={40 51 62 84} and (VX)=(XV)={03 12}. The quartet (ADVX)={4012 5103 6203 8403 8412} and the sextet (ACDVXZ)={490127 591037 692037 894037 894127}.
(WY)={16 27 38 49}, so (ACDVWXYZ)={49013287 89401367}.
The first option is missing the digits 5 and 6 which are adjacent, so satisfy the condition B=E+1, making E=5 and B=6.
The solution is therefore:
(ABCDEVWXYZ)=(4690513287).
Substituting into the original equations:
149+756=905, 481+392=873, therefore:
905-756=149, 873-392=481.
