How do I find the cost in a problem which doesn't mention the original number of elements? The problem is as follows:

Mike buys an unknown quantity of hardcover books for a total of
  $\textrm{328 USD}$. Later on, he realizes that had he bought their
  respective paperback versions they had cost $\textrm{4.5 USD}$ less
  each and by doing this he would have ended with $4$ more books and
  spend only $\textrm{320 USD}$. How much was the cost of each hardcover
  books which Mike bought?

This problem has made me to walk in circles for several hours and finally I thought that it has something to do with a quadratic equation or so.
What I thought was this:
If I want to relate the cost of something and the number of things I would use this formula:
$$x\,\left(\frac{\textrm{cost in whatever currency}}{\textrm{per item whatever is it}}\right ) \times\,\textrm{a elements}=\textrm{total cost}$$
Therefore I used this equation in my attempt to solve this problem:
I defined the number of hardcover books as $\textrm{H}$ and the number of paperback books as $\textrm{P}$, the cost for each of the hardcover books $\textrm{x}$ and the cost for each of the paperback books as $x-4.5$ as in the problem states they would had cost $\textrm{4.5 USD}$ less each.
The first part of the problem does not specify how many and how much for the hardcover books just the total cost so the equation becomes into:
$$\left (x \right) \left ( \textrm{H}\right ) = 328$$
The second part of the problem does mention that the cost of the paperback books is $\textrm{320 USD}$ so it becomes into:
$$\left (x-4.5 \right) \left ( \textrm{P}\right ) = 320$$
There is a third relationship which has been mentioned in the problem and that is the number of paperback books is the number of hardcover books plus four therefore:
$$P=H+4$$
So the second equation becomes into:
$$\left (x-4.5 \right) \left ( \textrm{H+4}\right ) = 320$$
Then all that is left to do is solve the system:
$$\left (x \right) \times \left ( \textrm{H}\right ) = 328$$
$$\left (x-4.5 \right) \times \left ( \textrm{H+4}\right ) = 320$$
Therefore:
$$xH+4x-4.5H-4.5\times4=320$$
inserting into the previous equation the first one and $$H=\frac{328}{x}$$
$$328+4x-4.5\times\left( \frac{328}{x}\right )-18=320$$
$$4x-\frac{1476}{x}-10=0$$ 
Multiplying by $x$ in both sides:
$$4x^{2}-10x-1476=0$$
Simplifying:
$$2x^{2}-5x-738=0$$
To solve the quadratic equation I used the formula:
$$X_{1,2}=\frac{5 \pm \sqrt{5^{2}-4\times\left( 2 \times \left(-738 \right)\right ) }}{2\times 2}$$
$$x_{1,2}=\frac{5 \pm \sqrt{5929}}{4}$$
Then the answers become
$$x_{1}=\frac{5+77}{4}=\frac{82}{4}=20.5$$
$$x_{2}=\frac{5-77}{4}=\frac{-72}{4}=-18$$
Now it is obvious that I cannot choose the negative value and since I defined as $x$ to be cost of each hardcover book the answer must be $\textrm{20.5 USD}$.
And the cost of the paperback book each must be:
$$20.5-4.5=16.0\,\textrm{USD}$$
althought the latter was not asked.
Is what I did correct?. The answer seems reasonable, but I'm not very sure if my substitutions are correct.
Is this the only way to approach this problem or could be another way to solve it faster?.
 A: cost per book initial - cost per book final $ =4.5$ 
$\dfrac{328}{x}-\dfrac{320}{x+4}=\dfrac{9}{2}\tag{1}$
$$9x^2+20x-2624=0$$
$$x=16 \ books$$
so cost of each hardcover book would be : $\dfrac{328}{16}=20.5 $USD   
edit: your approach is correct too..but i think  without introducing H (by thinking in terms of cost per book) you can save your time.
A: Assume that Mike bought $n$ harcover books at $h$ dollars per book. This gives the equation
$$n\cdot h=328\ .\tag{1}$$
The sequel of the story then tells us that
$$(n+4)\left(h-{9\over2}\right)=320\ ,\tag{2}$$
so that subtracting $(1)$ from $(2)$ we obtain the linear equation
$$4h-{9\over2}n-18=-8\ ,$$
or $n={2\over 9}(4h-10)$. Plugging this into $(1)$ leads to the quadratic equation
$$4h^2-10h-1476=0$$
with the single positive solution $h=20.5$. This leads to $n=16$, and it is easily verified that therewith all conditions of the story are satisfied.
A: Number of hardcovers = $n$
Cost of 1 hardcover $(c)$ = $\frac{328}{n}$
$\frac{328}{n}$ is the cost of one hardcover so if the paperback is 4.5 USD cheaper then one paperback is $\frac{328}{n}-4.5$.
We then know that Mike can get 4 more paperbacks so the number of paperbacks he can buy is $n+4$. The number of paperbacks multiplied by the cost of each paperback is given as 320 thus:
$$(\frac{328}{n}-4.5)(n+4)=320$$
$$328+\frac{1312}{n}-4.5n-18=320$$
$$\frac{1312}{n}-4.5n=10$$
$$4.5n^2+10n-1312=0$$
$$9n^2+20n-2624=0$$
$$n=16,n=-\frac{164}{9}$$
$n$ must be positive (since it is the number of books), therefore $n=16$
Cost of 1 hardcover $(c)$ = $\frac{328}{n}$
So $C = \frac{328}{16}=20.50$ USD
So yes, you are indeed correct but I think you made it a bit too complicated.
