SOLUTION: "We are comparing two plotters. An HP and a Canon. The HP lists for $3427 and the Canon for $4995. A black cartridge for the HP is $56, The black for Canon is $81. How many cartrid
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Question 823785: "We are comparing two plotters. An HP and a Canon. The HP lists for $3427 and the Canon for $4995. A black cartridge for the HP is $56, The black for Canon is $81. How many cartridges for each one will be needed before the price comes out the same?"
This is a real-life application of the algebra I learned in school, and I can probably solve it on my own once I figure out how to set it up. This is what I have tried:
3427+56x=4995+81y
This just comes out to the solution 4994=4995 and I don't really understand why. Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website! ---
x = number of cartridges
y = cost of operation
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canon:
y = 81x + 4995
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hp:
y = 56x + 3427
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y = 81x + 4995
y = 56x + 3427
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put the system of linear equations into standard form
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81x - y = -4995
56x - y = -3427
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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answer:
x= -62.72
y= -85.32
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the "cost of operation" for both plotters is the same but "negative" ($-85.32) when you purchase a "negative" number of cartridges (-62.72) !
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that sounds strange, but there's a simple explanation.
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regardless how many cartridges you purchase for either plotter, the Canon will always be more expensive to operate than the HP.
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the graphs of the cost lines of the two plotters make that very clear.
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notice that the lines intersect at x= -62.72 y= -85.32, but the lines diverge for x > 0
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graph:
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Solve systems of linear equations up to 6-equations 6-variables:
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