SOLUTION: A bookstore can purchase several calculators for a total cost of $160. If each calculator cost $4 less, the bookstore could purchase 2 additional calculators at the same total cost

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Question 1116968: A bookstore can purchase several calculators for a total cost of $160. If each calculator cost $4 less, the bookstore could purchase 2 additional calculators at the same total cost. How many calculators can be purchased at the regular price?
Found 2 solutions by ankor@dixie-net.com, josgarithmetic:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A bookstore can purchase several calculators for a total cost of $160.
If each calculator cost $4 less, the bookstore could purchase 2 additional calculators at the same total cost.
How many calculators can be purchased at the regular price?
:
let c = no. of calc purchased at regular price
then
160%2Fc = original price per calc
:
160%2Fc - 4 = 160%2F%28%28c%2B2%29%29
multiply by c(c+2), cancel denominators
160(c+2) - 4c(c+2) = 160c
160c + 320 - 4c^2 - 8c = 160c
subtract 160c to both sides
320 - 4c^2 - 8c = 0
0 = 4c^2 + 8c - 320
simplify, divide by 4
c^2 + 2c - 80 = 0
factors to
(c+10)(c-8) = 0
positive solution
c = 8 calculators at regular price

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x calculators at the regular price;
x+2 calculators at the new price;

160%2Fx-----regular price

160%2F%28x%2B2%29------new price

New price is $4 less per calculator.
highlight_green%28160%2Fx-160%2F%28x%2B2%29=4%29
This is the equation to solve.
--
40%2Fx-40%2F%28x%2B2%29=1
%2840%28x%2B2%29-40%29%2F%28x%28x%2B2%29%29=1
40x%2B80-40x=x%5E2%2B2x
x%5E2%2B2x-80=0
%28x-8%29%28x%2B10%29=0
highlight%28x=8%29