What is the value of $|A \cap B|$?

Question

For $A$ and $B$ sets,

$|B \times (A \cup B)| = 108$, $|A - B| = 3$ and $|B-A| = 5$

What is the value of $|A \cap B|$?

I see that there's cartesian product. What should I think?

Answer 1

Let $x =|A\cap B|$, then $|A| = x+3$, $|B|= 5+x$ and $|A\cup B| = x+8$.

Now solve $$(x+5)(x+8) =|B|\cdot |A\cup B| = |B\times (A\cup B)|=108$$

Since $0=x^2+13x-68 = (x+17)(x-4)$ we get the solution $x=4.$

Answer 2

There is known that $\vert A \cup B \vert = \vert A \vert + \vert B \vert - \vert A \cap B \vert$.

You need to understand that $\vert A \vert = \vert A \cap B \vert + \vert A-B \vert \;$ and $\;\vert B \vert = \vert A \cap B \vert + \vert B-A \vert$.

Then, you can use the procedure that John Watson uses in his answer.