document.write( "Question 164690: The cashier in a movie box office sold 200 more adult admission tickets at $11.00 each than children admission tickets at $8.00 each. What is the minimum number of each type of ticket that the cashier can sell for the total receipts to be at least $5000. \n" ); document.write( "
Algebra.Com's Answer #121361 by mangopeeler07(462)\"\" \"About 
You can put this solution on YOUR website!
200 more adult admission tickets...than children admission tickets:\r
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\n" ); document.write( "\n" ); document.write( "a=c+200\r
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\n" ); document.write( "\n" ); document.write( "adults=$11 each
\n" ); document.write( "children=$8 each
\n" ); document.write( "total receipts=at least $5000\r
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\n" ); document.write( "\n" ); document.write( "\"11a%2B8c%3E=5000\"\r
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\n" ); document.write( "\n" ); document.write( "System:
\n" ); document.write( "a=c+200
\n" ); document.write( "\"11a%2B8c%3E=5000\"\r
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\n" ); document.write( "\n" ); document.write( "Use substitution
\n" ); document.write( "\"11%28c%2B200%29%2B8c%3E=5000\"\r
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\n" ); document.write( "\n" ); document.write( "Distribute
\n" ); document.write( "\"11c%2B2200%2B8c%3E=5000\"\r
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\n" ); document.write( "\n" ); document.write( "Combine like terms
\n" ); document.write( "\"19c%2B2200%3E=5000\"\r
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\n" ); document.write( "\n" ); document.write( "Subtract 2200 from both sides
\n" ); document.write( "\"19c%3E=2800\"\r
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\n" ); document.write( "\n" ); document.write( "Divide both sides by 19
\n" ); document.write( "\"c%3E=148\"--------this is rounded up from 147.3, since you can't sell .3 of a ticket.\r
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\n" ); document.write( "\n" ); document.write( "a=c+200
\n" ); document.write( "a=148+200
\n" ); document.write( "a=348\r
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\n" ); document.write( "\n" ); document.write( "\"a%3E=348\"----since c might be more than 148.\r
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\n" ); document.write( "\n" ); document.write( "Answer:
\n" ); document.write( "148 childrens tickets
\n" ); document.write( "348 adults tickets
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