document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=564240>Question 564240</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do I 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( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #365288 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=365288\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  How do I 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( "1st: 2x-1<BR>\n" );
document.write( "2nd: 2x+1<BR>\n" );
document.write( "3rd: 2x+3<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "2(2x+1)-(2x+3) = 2x-1+11<BR>\n" );
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document.write( "4x+2-2x-3 = 2x+10<BR>\n" );
document.write( "2x - 1 = 2x + 10<BR>\n" );
document.write( "-1 = 10<BR>\n" );
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document.write( "The conditions are contradictory.<BR>\n" );
document.write( "No solution.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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