Question 1120967
student plans to spend 200 on p notebooks.
if the price goes up by 10, the student can buy p - 1 notebooks.


let x = the price of one notebook.


you have two equations that need to be solved simultaneously.


they are:


x * p = 200
(x+10) * (p-1) = 200


solve for x in the first equation to get x = 200/p


replace x in the second equation by 200/p to get:


(x+10) * (p-1) = 200 becomes:


(200/p + 10) * (p-1) = 200


multiply both sides of this equation by p to get:


p * (200/p + 10) * (p-1) = 200 * p


since p * (200/p + 10) is equal to 200 + 10p, the equation becomes:


(200 + 10p) * (p-1) = 200p


simplify to get:


200p - 200 + 10p * p - 10p = 200p


the 200p in both sides of the equation cancels out and you are left with:


-200 + 10p^2 - 10p = 0


rearrange the terms in descending order of degree to get:


10p^2 - 10p - 200 = 0


divide both sides of the equation by 10 to get:


p^2 - p - 20 = 0


factor this quadratic equation to get:


(p-5)*(p+4) = 0


solve for p to get p = 5 or p = -4


your solution is p = 5.


when p = 5, x * p = 200 becomes x * 5 = 200 which makes x = 40


when p = 4, x * p = 200 becomes x * 4 = 200 which makes x = 50


x is the price of each notebook.


when the price is 40, you can buy 200 / 40 = 5 notebooks.


when the price is raised by 10 to be equal to 50, you can buy 200 / 50 = 4 notebooks which is 1 less than when the price was 40.


your solution is that the price of each notebook was 40 before the increase.