document.write( "Question 143084: A chicken farmer and his wife are having a discussion about the future of their farm. The husband says, \"I think we should sell off 75 chickens, then we'd have enough feed for 20 days longer than it will last now.\" His wife, however, wants them to buy 100 more chickens, but he reminds her, \"Dear, that will mean that we'll run out of feed 15 days sooner than it will last now.\"\r
\n" ); document.write( "\n" ); document.write( "How many chickens do they have right now, and how many days will their feed last?
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Algebra.Com's Answer #104281 by ankor@dixie-net.com(22740)\"\" \"About 
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A chicken farmer and his wife are having a discussion about the future of their farm. The husband says, \"I think we should sell off 75 chickens, then we'd have enough feed for 20 days longer than it will last now.\" His wife, however, wants them to buy 100 more chickens, but he reminds her, \"Dear, that will mean that we'll run out of feed 15 days sooner than it will last now.\"
\n" ); document.write( "How many chickens do they have right now, and how many days will their feed last?
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\n" ); document.write( "Let x = no. of chicks now
\n" ); document.write( "Let y = no. of days of feed for x chicks
\n" ); document.write( "then
\n" ); document.write( "Total amt of feed = xy; call it \"chick-days\" or something like that
\n" ); document.write( ":
\n" ); document.write( "\"sell off 75 chickens, then we'd have enough feed for 20 days longer than it will last now.:
\n" ); document.write( "(x-75)(y+20) = xy
\n" ); document.write( "xy +20x - 75y - 1500 = xy
\n" ); document.write( "xy - xy + 20x - 75y = 1500
\n" ); document.write( "20x - 75y = 1500
\n" ); document.write( ":
\n" ); document.write( "\"buy 100 more chickens, we'll run out of feed 15 days sooner than it will last now.\"
\n" ); document.write( "(x+100)(y-15) = xy
\n" ); document.write( "xy - 15x + 100y - 1500 = xy
\n" ); document.write( "xy - xy - 15x + 100y = 1500
\n" ); document.write( "-15x + 100y = 1500
\n" ); document.write( ":
\n" ); document.write( "Two manageable equations:
\n" ); document.write( "Multiply the 1st equation by 3 and the 2nd equation by 4; add
\n" ); document.write( "60x - 225y = 4500
\n" ); document.write( "-60x +400y = 6000
\n" ); document.write( "---------------------adding eliminates x
\n" ); document.write( "0x + 175y = 10500
\n" ); document.write( "y = \"10500%2F175\"
\n" ); document.write( "y = 60 days of feed originally
\n" ); document.write( ":
\n" ); document.write( "Find x using 1st equation:
\n" ); document.write( "20x - 75(60) = 1500
\n" ); document.write( "20x - 4500 = 1500
\n" ); document.write( "20x = 1500 + 4500
\n" ); document.write( "x = \"6000%2F20\"
\n" ); document.write( "x = 300 chicks
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\n" ); document.write( ":
\n" ); document.write( "Check solution: we can say we have 60*300 = 18000 chick-days of food available
\n" ); document.write( ":
\n" ); document.write( "\"sell off 75 chickens, then we'd have enough feed for 20 days longer than it will last now.
\n" ); document.write( "300 - 75 = 225 chicks
\n" ); document.write( "18000/225 = 80 days (20 more days)
\n" ); document.write( ":
\n" ); document.write( "\"buy 100 more chickens, we'll run out of feed 15 days sooner than it will last now.\"
\n" ); document.write( "300 + 100 = 400 chicks
\n" ); document.write( "18000/400 = 45 days (15 days fewer)
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\n" ); document.write( "Did this make sense? What do you think? Let me know.\r
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