document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1027476>Question 1027476</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Twice the sum of Jerry and David's age is equivalent to ten years less than three times Frank's age. David is 9 years younger than Jerry and Frank is 5 years older than Jerry. Write and solve an algebraic equation to find the age of all three men.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #642680 by </B><A HREF=/tutors/aboutme.mpl?userid=macston><B>macston(5194)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=macston><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/cgi-bin/show-face.mpl?user=macston><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=642680\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .<BR>\n" );
document.write( "J=Jerry's age; D=David's age=J-9; F=Frank's age=J+5<BR>\n" );
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document.write( "2(J+D)=3F-10<BR>\n" );
document.write( "2(J+(J-9))=3(J+5)-10<BR>\n" );
document.write( "2(2J-9)=3J+15-10<BR>\n" );
document.write( "4J-18=3J+5<BR>\n" );
document.write( "J=23<BR>\n" );
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document.write( "D=J-9=23-9=14<BR>\n" );
document.write( ".<BR>\n" );
document.write( "F=J+5=23+5=28<BR>\n" );
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document.write( "ANSWER: Jerry is 23, David is 14, and Frank is 28.<BR>\n" );
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document.write( "CHECK:<BR>\n" );
document.write( "2(J+D)=3F-10<BR>\n" );
document.write( "2(23+14)=3(28)-10<BR>\n" );
document.write( "2(37)=84-10<BR>\n" );
document.write( "74=74<BR>\n" );
document.write( ".\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );