document.write( "Question 311293: at a fair the admission is $1.50 for children and $4.00 for adults. one day 2,200 people enter the fair and $5,050 is collected. how many children and how many adults attended. \n" ); document.write( "

Algebra.Com's Answer #222628 by fractalier(6550) $\"\"$ $\"About$

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Let x be the number of children.
\n" ); document.write( "Let y by the number of adults.\r
\n" ); document.write( "\n" ); document.write( "We have to set up two equations: One is for the number of people and the other is for the amount of money collected.\r
\n" ); document.write( "\n" ); document.write( "x + y = 2200
\n" ); document.write( "1.50x + 4.00y = 5050\r
\n" ); document.write( "\n" ); document.write( "We have to solve this system. I am going to do it by linear combination, that is, I am going to multiply the top equation by 4, so that, when I subtract the equations, the y's will drop out...here goes\r
\n" ); document.write( "\n" ); document.write( "4x + 4y = 8800
\n" ); document.write( "-(1.50x + 4.00y = 5050)\r
\n" ); document.write( "\n" ); document.write( "which yields\r
\n" ); document.write( "\n" ); document.write( "2.5x = 3750
\n" ); document.write( "and, then by dividing by 2.5, we get
\n" ); document.write( "x = 1500
\n" ); document.write( "Now substitute this back into the first equation, and
\n" ); document.write( "y = 700 \n" ); document.write( "

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