document.write( "Question 672545: A chemist has two alloys, one of which is 15% gold and 25% lead and the other which is 20% gold and 40% lead. How many grams of each of the two alloys should be used to make an alloy that contains 74 g of gold and 126 g of lead?
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Algebra.Com's Answer #418271 by ankor@dixie-net.com(22740)\"\" \"About 
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A chemist has two alloys, one of which is 15% gold and 25% lead and the other which is 20% gold and 40% lead.
\n" ); document.write( " How many grams of each of the two alloys should be used to make an alloy that contains 74 g of gold and 126 g of lead?
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\n" ); document.write( "The x = amt of the 1st alloy
\n" ); document.write( "let y = amt of the 2nd alloy
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\n" ); document.write( "Write a mixture equation for each amt:
\n" ); document.write( " gold percent
\n" ); document.write( ".15x + .20y = 74
\n" ); document.write( " the lead percent
\n" ); document.write( ".25x + .40y = 126
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\n" ); document.write( "multiply the 1st equation by 2, subtract the 2nd equation
\n" ); document.write( ".30x + .40y = 148
\n" ); document.write( ".25x + .40y = 126
\n" ); document.write( "--------------------subtraction eliminates y, find x
\n" ); document.write( ".05x = 22
\n" ); document.write( "x = \"22%2F.05\"
\n" ); document.write( "x = 440 grams of the 1st alloy
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\n" ); document.write( "Find y using the 1st original equation (gold)
\n" ); document.write( ".15(440) + .20y = 74
\n" ); document.write( "66 + .20y = 74
\n" ); document.write( ".20y = 74 - 66
\n" ); document.write( ".20y = 8
\n" ); document.write( "y = 8/.2
\n" ); document.write( "y = 40 grams of the 2nd alloy
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\n" ); document.write( "See if that checks out in lead equation
\n" ); document.write( ".25(440) + .40(40) = 126
\n" ); document.write( "110 + 16 = 126
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\n" ); document.write( "We can see we need 440 grams of the 1st alloy and 40 grams of the 2nd
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