document.write( "Question 918208: A hotel has 150 rooms. A room with extras costs $80 a night, and a room w/o costs $60 a night. On a night that the hotel is full, the total revenue is $10,000. How many of each room does the hotel have? \n" ); document.write( "
Algebra.Com's Answer #556955 by richwmiller(17219)\"\" \"About 
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x+y=150,
\n" ); document.write( "80*x+60*y=10000
\n" ); document.write( "x=150-y
\n" ); document.write( "80*(150-y)+60*y=10000
\n" ); document.write( "12000-80y+60*y=10000
\n" ); document.write( "-20*y=-2000
\n" ); document.write( "y=100
\n" ); document.write( "x=150-y
\n" ); document.write( "x=50 rooms at $80 ($4000) and y=100 rooms at $60 ($6000)
\n" ); document.write( "check
\n" ); document.write( "50+100=150
\n" ); document.write( "80*x+60*y=10000
\n" ); document.write( "80*50+60*100=10000
\n" ); document.write( "4000+6000=10000
\n" ); document.write( "10000=10000
\n" ); document.write( "ok
\n" ); document.write( "codetick
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