SOLUTION: A bookstore can purchase several calculators for a total cost of $216. If each calculator cost $2 less, the bookstore could purchase 9 additional calculators at the same total cost

Question 316581: A bookstore can purchase several calculators for a total cost of $216. If each calculator cost $2 less, the bookstore could purchase 9 additional calculators at the same total cost. How many calculators, n, can be purchased at the regular price?
Found 2 solutions by solver91311, texttutoring:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Solve for

John

Answer by texttutoring(324) (Show Source): You can put this solution on YOUR website!
Let n = number of calculators
Let p = original price of calculators

You have two variables, so you need two equations:

Equation 1: 216 = np
Equation 2: 216 = (n+9)(p-2)

Isolate for n in Equation 1: n = 216/p

Now plug this value into Equation 2 and solve for p:

216 =(216/p + 9)(p-2)

Do FOIL:

216 = 216 + 9p -432/p -18
0 = 9p -432/p -18

Multiply everything by p:

0 = 9p^2 -18p -432
0 = 9(p^2 -2p -48)
0 = 9(p-8)(p+6)

So either p=8 or p=-6. We have to choose the positive value, because a negative price wouldn't make sense.

Now let's solve for n:
n=216/p
n=216/8
n=27

So 27 calculators can be purchased at the regular price of $8 each.
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