document.write( "Question 873917: Six apples and 1 pear cost $2.00. At the same store, 3 apples and 6 pears cost $2.10 Find the cost of one apple and the cost of one pear.\r
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Algebra.Com's Answer #527169 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
If \"Six apples and 1 pear cost $2.00\", then \"6a%2B1p+=+2\" where 'a' is the cost for a single apple and p is the cost for a single pear.\r
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\n" ); document.write( "\n" ); document.write( "If \"3 apples and 6 pears cost $2.10\", then \"3a%2B6p+=+2.10\"\r
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\n" ); document.write( "\n" ); document.write( "The two equations you'll use are\r
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\n" ); document.write( "\n" ); document.write( "\"6a%2B1p+=+2\"
\n" ); document.write( "\"3a%2B6p+=+2.10\"\r
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\n" ); document.write( "\n" ); document.write( "Solve for p in \"6a%2B1p+=+2\" to get \"p+=+2-6a\"\r
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\n" ); document.write( "\n" ); document.write( "Then plug this into the second equation to get \r
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\n" ); document.write( "\n" ); document.write( "\"3a%2B6p+=+2.10\"\r
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\n" ); document.write( "\n" ); document.write( "\"3a%2B6%282-6a%29+=+2.10\"\r
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\n" ); document.write( "\n" ); document.write( "Solving that equation for 'a' will give you the cost for a single apple. Once you know the value of 'a', plug it into \"p+=+2-6a\" to find the value of p (the cost for a single pear)
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