SOLUTION: A school planned to buy x calculators for atotal cost of 16,200. The supplier agreed to offer a discount of 60 per calculator. The school was then able to buy three more calculator

Question 479766: A school planned to buy x calculators for atotal cost of 16,200. The supplier agreed to offer a discount of 60 per calculator. The school was then able to buy three more calculators for the same ammount of money.write an expression interms of x and find the original price of each calculator.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
A school planned to buy x calculators for atotal cost of 16,200. The supplier agreed to offer a discount of 60 per calculator. The school was then able to buy three more calculators for the same ammount of money.write an expression interms of x and find the original price of each calculator.
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Let p = the original price
The formula for the original transaction is
16200 = xp [1]
Since the total costs are the same, the discounted transaction can be written as
16200 = (x+3)(p-60) [2]
Solve for x in [1] and substitute into [2]:
16200 = (16200/p + 3)(p - 60)
16200 = 16200 - 16200*60/p + 3p - 180
0 = -972000/p + 3p - 180
Multiply through by p:
3p^2 - 180p - 972000 = 0
p^2 - 60p - 324000 = 0
p = (60 +- sqrt(3600 + 4*324000))/2
Taking the positive solution, this gives p = 600
So the original cost of each calculator is $600 [expensive calculators]

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