document.write( "Question 169246: A confectioner has candy costing 55 cents and 75 cents per kilogram. If he wishes to make 1000 kilograms of a mixture and sell it for 60 cents per kilogram, how much of each does he need? \n" ); document.write( "

Algebra.Com's Answer #124789 by Mathtut(3670) $\"\"$ $\"About$

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x+y=1000: eq 1 where x is the amount of 55 cent mixture and y is the amount
\n" ); document.write( "of 75 cent mixture
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\n" ); document.write( ".55x+.75y=.6(1000)...eq 2\r
\n" ); document.write( "\n" ); document.write( "lets take the y value of eq 1 which is y=1000-x and plug it into eq 2.
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\n" ); document.write( ".55x+.75(1000-x)=600
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\n" ); document.write( ".55x+750-.75x=600
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\n" ); document.write( ".2x=150\r
\n" ); document.write( "\n" ); document.write( " $\"highlight%28x=750%29\"$ kg of 55 cent candy
\n" ); document.write( " $\"highlight%281000-750=250%29\"$ kg of 75 cent candy \n" ); document.write( "

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