document.write( "Question 107664This question is from textbook beggining algebra
\n" ); document.write( ": 27. How much of an alloy that is 20% copper should be mixed with 200ounces of an alloy that is 50% copper in order to get an alloy of 30% copper? \n" ); document.write( "
Algebra.Com's Answer #78440 by solver91311(24713)\"\" \"About 
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You want to know how much 20% copper alloy, so let that be x. We also don't know the total amount of 30% alloy, so let that be y.\r
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\n" ); document.write( "\n" ); document.write( "Now, let's figure out what we do know.\r
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\n" ); document.write( "\n" ); document.write( "The amount of 20% alloy plus the 200 oz. of 50% alloy must equal the amount of 30% alloy, so:\r
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\n" ); document.write( "\n" ); document.write( "Eq 1) \"x%2B200=y\"\r
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\n" ); document.write( "\n" ); document.write( "We also know that the amount of copper in the 20% alloy is 0.2x, the amount of copper in the 50% alloy is 50% of 200 oz, or 100 oz, and the amount of copper in the 30% alloy must be 0.3y. We also know that the copper in the 20% alloy plus the copper in the 50% alloy must equal the copper in the 30% alloy. Therefore:\r
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\n" ); document.write( "\n" ); document.write( "Eq 2) \"0.2x%2B100=0.3y\"\r
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\n" ); document.write( "\n" ); document.write( "Now it is a matter of solving two simultaneous linear equations in two variables. There are a number of methods, but this is the one I like. First re-arrange the equations to standard form, \"ax%2Bby%2Bc=0\". Or, in this case:\r
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\n" ); document.write( "\n" ); document.write( "\"x-y%2B200=0\" and \"0.2x-0.3y%2B100=0\"\r
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\n" ); document.write( "\n" ); document.write( "Then I multiply the 2nd equation by -5: \"-x%2B1.5y-500=0\"\r
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\n" ); document.write( "\n" ); document.write( "Then add the result to the first equation, term by term: \"0x%2B0.5y-300=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"0.5y=300\"
\n" ); document.write( "\"y=600\"\r
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\n" ); document.write( "\n" ); document.write( "Then since \"x%2B200=600\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=400\"ounces.\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "\"400%2B200=600\" and\r
\n" ); document.write( "\n" ); document.write( "\"0.2%2A400%2B0.5%2A200=0.3%2A600\"
\n" ); document.write( "\"80+%2B+100+=+180\" Check!\r
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