Old 10-3-2006, 09:32 PM   #1
randonmess604
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Default help with algebra.....

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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Old 10-3-2006, 09:34 PM   #2
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Default Re: help with algebra.....

algebra is not critical thinking

if you want math help go to http:/www.artofproblemsolving.com or something
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Old 10-4-2006, 12:28 AM   #3
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Default Re: help with algebra.....

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

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