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#1 |
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FFR Player
Join Date: Oct 2006
Posts: 1
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ok...i know that this might not be the best place to put this....but i need help with solving linear equations involving ratios
A school play ran for two nights, with audiences totalling 1390 adults and students. They paid $4285 for admission. One adult ticket cost $4.00 and one student ticket cost $2.50. The ratio of adults to students was 3 : 5 on the first night and 2 : 3 on the second night. How many students attended each night? thanks for the help if can give any |
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#2 |
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FFR Player
Join Date: Aug 2005
Location: awsome
Posts: 2,946
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algebra is not critical thinking
if you want math help go to http:/www.artofproblemsolving.com or something
__________________
hehe |
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#3 |
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is against custom titles
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x - #Adults, x1 - #adults on night 1, x2 - #adults on night 2
y - #students, y1 - #students on night 1, y2 - #students on night 2 x + y = 1390 -> y = 1390 - x 4x + 2.5y = 4285 4x + 3475 - 2.5x = 4285 1.5x = 810 x = 540 total adults y = 850 total students x1 + 5/3x1 + x2 + 3/2x2 = 1390 x1 + x2 = 540 -> x1 = 540 - x2 540 - x2 + 900 - 5/3x2 + x2 + 3/2x2 = 1390 1440 - 8/3x2 + 5/2x2 = 1390 50 = 1/6x2 x2 = 300 -> x1 = 240 y2 = 450 -> y1 = 400 750 + 640 = 1390 450/300 = 3/2 , 400/200 = 5/3 --Guido http://andy.mikee385.com |
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