document.write( "Question 1077610: I cannot figure out how to get how many tickets were sold for each priced ticket. \r
\n" ); document.write( "\n" ); document.write( "A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3343 tickets overall. It has sold 113 more $20 tickets than $10 tickets. The total sales for $62,350. How many tickets have been sold?
\n" ); document.write( "How many $10 tickets were sold?
\n" ); document.write( "How many $20 tickets were sold?
\n" ); document.write( "How many $30 tickets were sold?
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Algebra.Com's Answer #692065 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
x, how many $10 tickets
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\n" ); document.write( "z, how many $30 VIP tickets\r
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\n" ); document.write( "\n" ); document.write( "Follow the description.
\n" ); document.write( "\"system%28x%2By%2Bz=3343%2Cy-x=113%2C10x%2B20y%2B30z=62350%29\"\r
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\n" ); document.write( "\n" ); document.write( "Three linear equations in three unknown variables, so solve the system.
\n" ); document.write( "You would want to simplify the sales equation to \"x%2B2y%2B3z=6235\", and the second equation maybe as \"-x%2By=113\".\r
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\n" ); document.write( "\n" ); document.write( "Work with THIS system:
\n" ); document.write( "\"system%28x%2By%2Bz=3343%2C-x%2By=113%2Cx%2B2y%2B3z=6235%29\"\r
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\n" ); document.write( "\n" ); document.write( "One path to begin can be, add first and second equations to form a new equation; and add the second and third equations to form another new equation. This eliminates x, and your two new equations are in just the two variables, y and z.
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