document.write( "Question 667200: A total of 42 tons of two types of ore is to be added into a smelter. The first type contains 6% copper and the other contains 2.5% copper. Find the necessary amounts of each ore to produce 2 tons of copper.\r
\n" ); document.write( "\n" ); document.write( "I need two equations. Thank you so much \n" ); document.write( "
Algebra.Com's Answer #414865 by mananth(16946)\"\" \"About 
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x 6% copper ore
\n" ); document.write( "y 2.5% copper ore
\n" ); document.write( "Total tons of ore
\n" ); document.write( "1.00 x + 1.00 y = 42.00 .............1
\n" ); document.write( "Quantity of copper
\n" ); document.write( "6.00 x + 2.50 y = 200.00 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by -2.50
\n" ); document.write( "Multiply (2) by 1.00
\n" ); document.write( "-2.50 x \ -2.50 y = -105.00
\n" ); document.write( "6.00 x + 2.50 y = 200.00
\n" ); document.write( "Add the two equations
\n" ); document.write( "3.50 x = 95.00
\n" ); document.write( "/ 3.50
\n" ); document.write( "x = 27.14
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1.00 x + 1.00 y = 42.00
\n" ); document.write( "27.14 + y = 42.00
\n" ); document.write( " y = 42.00 -27
\n" ); document.write( " y = 14.86
\n" ); document.write( " y = 14.86
\n" ); document.write( " 27.14 6% copper ore
\n" ); document.write( " 14.86 2.5% copper ore
\n" ); document.write( "m.ananth@hotmail.ca
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