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Question 824282: A book store can purchase several calculators for a total cost of $120. If each calculator cost $1 less, the bookstore could purchase 10 additional calculators at the same total cost. How many calculators can be purchased at the regular price?
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
x = number of calculators
y = price
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xy = 120
(x + 10)(y - 1) = 120
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y = 120/x
xy - x + 10y - 10 = 120
x120/x - x + 10*120/x - 10 = 120
-x + 1200/x - 10 = 0
-xx + 1200 - 10x = 0
-xx - 10x + 1200 = 0
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the above quadratic equation is in standard form, with a=-1, b=-10, and c=1200
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-1 -10 1200
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
x = -40
x = 30
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a negative number of calculators doesn't make sense for this problem, so use the positive root:
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answer:
x = number of calculators = 30
y = price = $4
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Solve quadratic equations, quadratic formula:
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Solve systems of linear equations up to 6-equations 6-variables:
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