document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=172055>Question 172055</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find three consecutive odd integers such that three times the second minus the third is 11 more than the first.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #127093 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=127093\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find three consecutive odd integers such that three times the second minus the third is 11 more than the first. <BR>\n" );
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document.write( "They're x, x+2 and x+4<BR>\n" );
document.write( "Three times the 2nd is 3x+6<BR>\n" );
document.write( "3x+6 - (x+4) = x + 11<BR>\n" );
document.write( "3x #2...3rd....1st+11<BR>\n" );
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document.write( "3x+6 - (x+4) = x + 11<BR>\n" );
document.write( "2x + 2 = x + 11<BR>\n" );
document.write( "x = 9<BR>\n" );
document.write( "So they're 9, 11 and 13\r<BR>\n" );
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