document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=162804>Question 162804</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find 3 #'s in which the second # is 3 times the first # and the third # is 2 less than 5 times the first #.  The sum of three #'s is 115.  Find the three #'s.\r<BR>\n" );
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document.write( "What I have tried:\r<BR>\n" );
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document.write( "3(x+2)+2(5x+1)=115<BR>\n" );
document.write( "3x+6+10x+1=115<BR>\n" );
document.write( "13x+7=115<BR>\n" );
document.write( "   -7  -7<BR>\n" );
document.write( "13x=108<BR>\n" );
document.write( "/13 /13<BR>\n" );
document.write( "x=8.31\r<BR>\n" );
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document.write( "I know this isn't correct. Please help<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #119998 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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=119998\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> S=3F<BR>\n" );
document.write( "T=5F-2<BR>\n" );
document.write( "F+S+T=115<BR>\n" );
document.write( "F+3F+(5F-2)=115<BR>\n" );
document.write( "9F=115+2<BR>\n" );
document.write( "9F=117<BR>\n" );
document.write( "F=117/9<BR>\n" );
document.write( "F=13 FOR THE FIRST NUMBER.<BR>\n" );
document.write( "3*13=39 FOR THE SECOND NUMBER.<BR>\n" );
document.write( "5*13-2=65-2=63 FOR THE THIRD NUMBER.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( "13+39+63=115<BR>\n" );
document.write( "115=115 </SPAN>\n" );
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