The answer depends on whether you mean
A intersecting (B union C) or
(A intersecting B) union C.
If: a = b+c+d Then: c = a-b-d
not (b or c) = (not b) and (not c)
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
ax - b = c ax = b + c x = (b + c)/a
There is nothing to "solve". You can evaluate the expression when each of a, b and c are TRUE or FALSE. But that is not solving.
exactly one
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
If: a = 2b+c Then: a-c = 2b And: b = (a-c)/2
complement of c
c equals b over 8
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c
to solve ab=c+a for a:Divide both sides by b, so:a= (c+a)/b