For sets $A$ and $B$, $|A∪B|=45,|A|=30, \text{and} |A∩B|=7.$ Find $|B|$ For sets $A$ and $B$, $|A\cup B| = 45, |A| = 30$, and $|A\cap B|=7$.  Find $|B|$
To solve, I believe you take $|A\cup B| + |A\cap B|= 45 + 7$
Together that yields 52.  Then you subtract $|A| = 30$ from 52, which gives 22.  So I believe $|B| = 22$
If we take a simpler example, let $A$ = {1,2,3,4,5}, and $B$ = {3,4,5,6}.  In that case, $|A \cup B| = 6$.  $|A \cap B| = 3$.  6+3 = 9.  Take 9-5 since $|A|=5$, and indeed you come up with 4, which =$|B|$.
Is my reasoning correct?
 A: 
$ (A \cup B)= B \cup (A-B)$
as $B $ and $ (A-B)$ are disjointed
then 
$$ |A \cup B|= |B| + |A-B|$$
and we have 
$A = (A \cap B) \cup (A-B)$
as $(A \cap B)$ and $(A-B)$ are disjointed
then 
 $$|A| = |A \cap B| +|A-B|$$ 
so $$ |A \cup B|= |B| + |A-B| = |B| + (|A| - |A \cap B|)$$
so $$ |B| = |A \cup B|- (|A| - |A \cap B|)  $$
so $$|B| =22$$
A: Although there are already plenty of correct answers, I'd like to describe how I'd think about this problem, without writing a lot of formulas.  You're given that $A$ and $B$ together have 45 elements, and that 30 of these are in $A$. So the remaining 15, the ones that aren't in $A$, have to be in $B$.  In addition, you're given that $B$ contains 7 elements that are in $A$.  So $B$ has 15 elements that are not in $A$ and 7 more that are in $A$, making 22 altogether.
A: $$|A\cup B| = |A| + |B| - |A \cap B| \implies |B| = |A \cup B| + |A \cap B|  - |A|.$$
A: Yes, we have: $$|A \cup B| = |A|+|B| - |A \cap B|.$$When we count the elements in $A \cup B$, we count the ones in $A \cap B$ twice, hence the subtraction. So: $$45=30+|B| - 7 \implies |B| = 22,$$alright.
