document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=21115>Question 21115</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I've been trying to help my son with this problem but can't figure out where to start.  The problem is this, Twice the greater of two consecutive odd integers is 13 less than three times the lesser number.  Find the integers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #10147 by </B><A HREF=/tutors/aboutme.mpl?userid=suresh><B>suresh(20)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=suresh><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=10147\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the our required odd number.<BR>\n" );
document.write( "Since the two integers are consecutive, the other number is x+2.\r<BR>\n" );
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document.write( "Given that, 2(greater number)=3(lesser number)-13.\r<BR>\n" );
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document.write( "2(x+2)=3x-13<BR>\n" );
document.write( "2x+4=3x-13<BR>\n" );
document.write( "Subtract 2x and add 13 on both sides,<BR>\n" );
document.write( "2x+4-2x+13=3x-13-2x+13<BR>\n" );
document.write( "17=x\r<BR>\n" );
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document.write( "Therefore x = 17.\r<BR>\n" );
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document.write( "The other number is x+2 = 19.\r<BR>\n" );
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document.write( "Hence the intergers are 17 and 19. </SPAN>\n" );
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