document.write( "Question 1170750: A movie theater has a seating capacity of 299. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 2164, How many children, students, and adults attended?
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Algebra.Com's Answer #795630 by ikleyn(52781)\"\" \"About 
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\n" ); document.write( "A movie theater has a seating capacity of 299.
\n" ); document.write( "The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.
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document.write( "Let x be the number of adults.\r\n" );
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document.write( "Then the number of children is  2x\r\n" );
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document.write( "and the number of students is  (299-x-2x) = 299-3x  (assuming full capacity is occupied).\r\n" );
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document.write( "Next you write the money equation (= the revenue equation)\r\n" );
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document.write( "    12x + 5*(2x) + 7*(299-3x) = 2164   dollars\r\n" );
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document.write( "It is your basic (governing) equation for this problem.\r\n" );
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document.write( "Now simplify and solve it\r\n" );
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document.write( "    12x + 10x +7*299 - 21x = 2164\r\n" );
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document.write( "           x               = 2164 - 7*299 = 71.\r\n" );
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document.write( "ANSWER.  71 adults,  2*71 = 142 children  and the rest  299-71-142 = 86 are students.\r\n" );
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document.write( "CHECK.   71*12 + 142*5 + 86*7 = 2164 dollars is the total revenue.    ! Correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "The amazing unexpected fact is that the problem is solved using one single unknown and one equation.\r
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