Question 1107938
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<pre>
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 = {{{36000/2400}}} = 15.


Then from eq(2''')  10N = 16000 + 1600*15 = 40000  ====>  N = 4000.


<U>Answer</U>.  N= 4000,  p= 15.
</pre>

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To see many other similar solved problems, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Had-they-sold.lesson>Had they sold . . .</A>

in this site.



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&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Miscellaneous word problems</U>",

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


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