document.write( "Question 414458: A $3.00 toll is charged to cross the bridge from Sanibel Island to mainland Florida. A six-month pass, costing $15.00, reduces the toll to $.50. A one-year pass, costing $150, allows for free crossings. How many crossings per year does it take, on average, for two six-months passes to be the most economical choice? Assume a constant number of trips per month, Show work and explain answer. \r
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Algebra.Com's Answer #290723 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"x\" = crossings per year it takes, on average, for two six-months passes to be the
\n" ); document.write( "most economical choice.
\n" ); document.write( "Let \"C\" = cost per year to make \"x\" crossings
\n" ); document.write( "Assume a constant number of trips per month
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\n" ); document.write( "Without buying any passes:
\n" ); document.write( "\"C%5B1%5D+=+3x\"
\n" ); document.write( "Buying 2 6-month passes:
\n" ); document.write( "\"C%5B2%5D+=+.5x+%2B+2%2A15\"
\n" ); document.write( "Buying a 1-year pass:
\n" ); document.write( "\"C%5B3%5D+=+150\"
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\n" ); document.write( "First I'll set \"C%5B1%5D+=+C%5B2%5D\" to find intersection
\n" ); document.write( "\"3x+=+.5x+%2B+2%2A15\"
\n" ); document.write( "\"3x+=+.5x+%2B+30\"
\n" ); document.write( "\"2.5x+=+30\"
\n" ); document.write( "\"x+=+12\"
\n" ); document.write( "12 crossings per year make \"C%5B1%5D+=+C%5B2%5D\"
\n" ); document.write( "One more crossing, \"x+=+13\", make
\n" ); document.write( "\"C%5B1%5D+=+3%2A13\"
\n" ); document.write( "\"C%5B1%5D+=+39\"
\n" ); document.write( "and
\n" ); document.write( "\"C%5B2%5D+=+.5%2A13+%2B+30\"
\n" ); document.write( "\"C%5B2%5D+=+36.5\"
\n" ); document.write( "So, \"C%5B2%5D\" becomes more economical on the 13th crossing by $2.50
\n" ); document.write( "and it still cost less than the 1-year pass for $150
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