The way this question is worded suggests that the question marks and the variables a, b, c, d each represent a different integer between 1 and 12, because there are four equations and one variable and two question marks on each line.
However, there are no solutions possible under this assumption for the following reasons:
Assume the second line means ?×b=?, so that in the four lines we have all four operations: add, multiply, subtract, divide.
Consider just multiply and divide. If we write all possible combinations of integers between 1 and 12, we get (2,3,6) because 2×3=6, 6/2=3, etc. (2,4,8), (2,5,10), (2,6,12), (3,4,12). We need a pair of these groups of 3 such that no integers are repeated, so the only pair is (2,5,10) and (3,4,12). This pair would be involved in line 2 and line 4. This leaves us with the integers 1, 6, 7, 8, 9, 11 for the addition and subtraction. This gives us the additions 1+6=7, 1+7=8, 1+8=9 only. Each one of these uses 1 so there is no way to find a pair of these without duplicating 1.
Now let’s assume the second line ?b=? means we place another single digit in front of a digit represented by b. That gives us only one solution: 12=12 where b=2, without duplicating any integers. That leaves us with the remaining integers for the other three lines: 3, 4, 5, 6, 7, 8, 9, 10, 11. With these integers we can’t satisfy line 4, because we need a 2 so that we can use 6/3=2, 8/4=2, 10/5=2.
So there are no solutions under theses assumptions.
We can choose 8 integers in the range 1-12 and allow a, b, c, d to repeat one of these. So let’s make such an assumption.
Here is one solution:
Line 2: 2×5=10, so b=5 or 2.
Line 4: 12/3=4 or 12/4=3, so b=12.
Or:
Line 2: 3×4=12, so b=3 or 4.
Line 4: 10/5=2 or 10/2=5, so b=10.
Now addition and subtraction.
Put a=1, then 6+a=7 for line 1.
Line 3 can be 9-8=1, c=9.
6+a=7, a=1
2b=10, b=5
c-8=1, c=9
d/3=4, d=12