document.write( "Question 21082: THE SUM OF A NUMBER AND ITS RECIPORCAL IS 10 OVER 3 WHAT IS THAT NUMBER? USE FIVE STEP METHOD I NEED MY DIPLOMA!! HELP!~! \n" ); document.write( "
Algebra.Com's Answer #10132 by Earlsdon(6294)\"\" \"About 
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I'm not sure what the \"five-step\" method is but perhaps you can adapt the following to fit. Let n be the unknown number.\r
\n" ); document.write( "\n" ); document.write( "Write the equation:
\n" ); document.write( "\"n+%2B+1%2Fn+=+10%2F3\" Add the terms on the left side.
\n" ); document.write( "Solve the equation:
\n" ); document.write( "\"%28n%5E2+%2B+1%29%2Fn+=+10%2F3\" Multiply both sides by n.
\n" ); document.write( "\"n%5E2+%2B+1+=+10n%2F3\" Multiply both sides by 3.
\n" ); document.write( "\"3n%5E2+%2B+3+=+10n\"} Subtract 10n from both sides.
\n" ); document.write( "\"3n%5E2+-+10n+%2B+3+=+0\" Solve this quadratic equation for n by factoring.
\n" ); document.write( "\"%283n+-+1%29%28n+-+3%29+=+0\" Apply the zero products principle.
\n" ); document.write( "\"3n+-+1+=+0\" and/or \"n+-+1+=+0\"
\n" ); document.write( "If \"3n+-+1+=+0\", then \"3n+=+1\" and \"n+=+1%2F3\"
\n" ); document.write( "If \"n+-+3+=+0\", then \"n+=+3\"\r
\n" ); document.write( "\n" ); document.write( "So there are two answers to this problem:
\n" ); document.write( "n = 3 and n = 1/3\r
\n" ); document.write( "\n" ); document.write( "Check:\r
\n" ); document.write( "\n" ); document.write( "n = 3: \"3+%2B+1%2F3+=+9%2F3+%2B+1%2F3\" = \"10%2F3\" OK
\n" ); document.write( "n = 1/3: \"1%2F3+%2B+1%2F%281%2F3%29+=+1%2F3+%2B+3\" = \"1%2F3+%2B+9%2F3+=+10%2F3\" OK\r
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