.
Let N be the number of copies and p be the price for one single copy in the base scenario.
Then the condition gives you this system of 2 non-linear equations
Np = (N+2000)*(p-5), (1)
Np = (N-1600)*(p+10), (2)
It is equivalent to
Np = Np + 2000p - 5N - 10000, (1')
Np = Np - 1600p + 10N - 16000. (2')
Simplifying it, you get
2000p - 5N = 10000, (1'')
-1600p + 10N = 16000. (2'')
This system is just a linear, and you can easily solve it by using the Elimination method:
4000p - 10N = 20000 (1''')
-1600p + 10N = 16000 (2''')
-----------------------------Add the equations
2400p = 36000 ====> p = = 15.
Then from eq(2''') 10N = 16000 + 1600*15 = 40000 ====> N = 4000.
Answer. N= 4000, p= 15.
--------------
To see many other similar solved problems, look into the lesson
- Had they sold . . .
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Miscellaneous word problems",
where you can find tons of interesting problems.
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.