document.write( "Question 524212: An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
\n" ); document.write( "water. Assuming that the silver loses 1/10 of its weight in water and that
\n" ); document.write( "gold loses 1/19 of its weight, how many ounces of each metal are in the
\n" ); document.write( "alloy?
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Algebra.Com's Answer #347718 by ankor@dixie-net.com(22740)\"\" \"About 
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An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
\n" ); document.write( "water. Assuming that the silver loses 1/10 of its weight in water and that
\n" ); document.write( "gold loses 1/19 of its weight, how many ounces of each metal are in the
\n" ); document.write( "alloy?
\n" ); document.write( ":
\n" ); document.write( "Let s = amt of silver
\n" ); document.write( "Let g = amt of gold
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\n" ); document.write( "\"An alloy of silver and gold weighs 15 ounces in air\"
\n" ); document.write( "s + g = 15
\n" ); document.write( "s = (15-g)
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\n" ); document.write( "And 14 ounces in water.
\n" ); document.write( ": Assuming that the silver loses 1/10 of its weight in water and that
\n" ); document.write( "gold loses 1/19 of its weight,
\n" ); document.write( "therefore
\n" ); document.write( "\"9%2F10\"s + \"18%2F19\"g = 14
\n" ); document.write( "Multiply by 190, to get rid of the denominators
\n" ); document.write( "19(9s) + 10(18g) = 190(14)
\n" ); document.write( "171s + 180g = 2660
\n" ); document.write( ":
\n" ); document.write( "replace s with (15-g), find g:
\n" ); document.write( "171(15-g) + 180g = 2660
\n" ); document.write( "2565 - 171g + 180g = 2660
\n" ); document.write( "-171g + 180g = 2660 - 2565
\n" ); document.write( "9g = 95
\n" ); document.write( "g = \"95%2F9\"
\n" ); document.write( "g = 10.556 oz of gold
\n" ); document.write( "then
\n" ); document.write( "15 - 10.556 = 4.444 oz of silver\r
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