document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=27432>Question 27432</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need a little help with this word problem.  I have tried to fiugre out how to write it out into algebraic  form...but have been unsuccessful.  Any help would be appreciated.<BR>\n" );
document.write( "The inscription on Greek Diophantus' tomb:\r<BR>\n" );
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document.write( "One-sixth of his life God granted him youth. After a twelfth more, he grew a beard.  After an additional seventh, he married, and 5 years later, he had a son.  Alas, the unfortunate son's life span was only one-half that of his father, who consoled his grief in the remaining 4 years of his lfe.\r<BR>\n" );
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document.write( "Use your knowledge to find how many years Diophantus lived. (HINT: let x be the number of years Diophantus lived.)\r<BR>\n" );
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document.write( "Again thanks for the help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #14894 by </B><A HREF=/tutors/aboutme.mpl?userid=Paul><B>Paul(988)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Paul><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=14894\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be his age:<BR>\n" );
document.write( "After = + and 5 years after means that add 5 to current status:<BR>\n" );
document.write( "1x/6+1x/12+1x/7+5+4=1x/2<BR>\n" );
document.write( "Common denominator = 84<BR>\n" );
document.write( "14x+7x+12x+420+336=42x<BR>\n" );
document.write( "Simplfy<BR>\n" );
document.write( "33x+756=42x<BR>\n" );
document.write( "9x=756<BR>\n" );
document.write( "x=84\r<BR>\n" );
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document.write( "Hence, he lived for 84 years.<BR>\n" );
document.write( "Paul. </SPAN>\n" );
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