document.write( "Question 250801: A mixture of 30 pounds of candy sells for $1.10 a pound. The mixture consists of chocolates worth $1.50 a pound and chocolates worth 90¢ a pound. How many pounds of the $1.50 chocolate were used to make the mixture?\r
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document.write( "A)15
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document.write( "B)10
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document.write( "C)12
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Algebra.Com's Answer #184565 by Theo(13342)![]() ![]() You can put this solution on YOUR website! your equations are \n" ); document.write( "1.5*x + .9*y = 1.10*30 \n" ); document.write( "x + y = 30 \n" ); document.write( "solve for x in the second equation: \n" ); document.write( "x = 30-y \n" ); document.write( "substitute for x in the first equation: \n" ); document.write( "1.5*(30-y) + .9*y = 1.10*30 \n" ); document.write( "simplify by removing parentheses: \n" ); document.write( "1.5*30 - 1.5*y + .9*y = 1.10*30 \n" ); document.write( "subtract 1.5*30 from both sides of the equation: \n" ); document.write( "-.6*y = 1.10*30 - 1.5*30 \n" ); document.write( "combine like terms: \n" ); document.write( "-.6*y = -.4*30 \n" ); document.write( "solve for y: \n" ); document.write( "y = 20\r \n" ); document.write( "\n" ); document.write( "confirm by substituting in original equation: \n" ); document.write( "x = 10 \n" ); document.write( "y = 20 \n" ); document.write( "1.5*10 + .9*20 = 1.10*30 \n" ); document.write( "simplify: \n" ); document.write( "15 + 18 = 33 \n" ); document.write( "confirm: \n" ); document.write( "33 = 33\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "answer is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "10 pounds of the $1.50 chocolate were used. \n" ); document.write( " \n" ); document.write( " |