document.write( "Question 770163: solve application problem- system of linear equations in 3 variables\r
\n" ); document.write( "\n" ); document.write( "x=children $3
\n" ); document.write( "y=students $4
\n" ); document.write( "z= adult $5\r
\n" ); document.write( "\n" ); document.write( " 1,000 tickets were sold for a play, which generated $3,800 in revenue. The ticket prices were $3 for children, $4 for students, & $5 for adults. There were 100 fewer student tickets sold than adult tickets. Find the number of each type of ticket sold.\r
\n" ); document.write( "\n" ); document.write( "x+y+z=1,000
\n" ); document.write( "3x+4y+5z=3,800\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "y=z-100
\n" ); document.write( "y-z=-100
\n" ); document.write( "

Algebra.Com's Answer #469332 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
You have all the right equations. At least try a substitution. Eliminate z through substitution into the first two equations, and then solve for x and y.\r
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\n" ); document.write( "\n" ); document.write( "Tickets: x+y+z=1,000
\n" ); document.write( "Money: 3x+4y+5z=3,800
\n" ); document.write( "Adults versus students: z=y+100\r
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\n" ); document.write( "\n" ); document.write( "\"x%2By%2By%2B100=1000\" and \"3x%2B4y%2B5%28y%2B100%29=3800\"
\n" ); document.write( "\"x%2B2y=900\" and \"3x%2B9y=3300\"
\n" ); document.write( "\"x%2B2y=900\" and \"x%2B3y=1100\"\r
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\n" ); document.write( "\n" ); document.write( "Add the opposite of the 900 equation to the 1100 equation.
\n" ); document.write( "\"x%2B3y-x-2y=1100-900\"
\n" ); document.write( "\"highlight%28y=200%29\",
\n" ); document.write( "So, what is x?
\n" ); document.write( "and then what is z?
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