document.write( "Question 770163: solve application problem- system of linear equations in 3 variables\r
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document.write( "x=children $3
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document.write( "y=students $4
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document.write( "z= adult $5\r
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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
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document.write( "x+y+z=1,000
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document.write( "3x+4y+5z=3,800\r
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document.write( "y=z-100
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document.write( "y-z=-100 \n" );
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Algebra.Com's Answer #469332 by josgarithmetic(39618)![]() ![]() ![]() 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 \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Add the opposite of the 900 equation to the 1100 equation. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So, what is x? \n" ); document.write( "and then what is z? \n" ); document.write( " |