document.write( "Question 284735: Please help me solve this equation:
\n" ); document.write( "\"3x+%2B+6y+=+48\"
\n" ); document.write( "\"-5x+%2B+6y+=+32\"
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Algebra.Com's Answer #206560 by Theo(13342)\"\" \"About 
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first equation is 3x + 6y = 48
\n" ); document.write( "second equation is -5x + 6y = 32\r
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\n" ); document.write( "\n" ); document.write( "subtract second equation from first equation to get:\r
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\n" ); document.write( "\n" ); document.write( "8x = 16\r
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\n" ); document.write( "\n" ); document.write( "divide both sides of this equation by 8 to get:\r
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\n" ); document.write( "\n" ); document.write( "x = 2\r
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\n" ); document.write( "\n" ); document.write( "replace x with 2 in the first equation to get:\r
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\n" ); document.write( "\n" ); document.write( "6 + 6y = 48\r
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\n" ); document.write( "\n" ); document.write( "subtract 6 from both sides of this equation to get:\r
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\n" ); document.write( "\n" ); document.write( "6y = 42\r
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\n" ); document.write( "\n" ); document.write( "divide both sides of this equation by 6 to get:\r
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\n" ); document.write( "\n" ); document.write( "y = 7\r
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\n" ); document.write( "\n" ); document.write( "your answer should be x = 6 and y = 7.\r
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\n" ); document.write( "\n" ); document.write( "substitute these values in both original equations to get:\r
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\n" ); document.write( "\n" ); document.write( "first equation is 3x + 6y = 48
\n" ); document.write( "second equation is -5x + 6y = 32\r
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\n" ); document.write( "\n" ); document.write( "3x + 6y becomes 3*2 + 6*7 = 6 + 42 = 48 confirming x and y are good in the first equation.\r
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\n" ); document.write( "\n" ); document.write( "-5x + 6y = 32 becomes -5*2 + 6*7 becomes -10 + 42 = 32 confirming x and y are good in the second equation.\r
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\n" ); document.write( "\n" ); document.write( "your answer is:\r
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\n" ); document.write( "\n" ); document.write( "x = 2 and y = 7\r
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