document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=210394>Question 210394</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 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 #159009 by </B><A HREF=/tutors/aboutme.mpl?userid=MathTherapy><B>MathTherapy(10555)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=MathTherapy><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=MathTherapy><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=159009\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 more than the first. \r<BR>\n" );
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document.write( "Let the 1st integer be F, then the 2nd is F + 2, and the 3rd is F + 4\r<BR>\n" );
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document.write( "Since 3 times the 2nd minus the 3rd is 31 more than the first, we'll have:\r<BR>\n" );
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document.write( "3(F + 2) - (F + 4) = F + 31\r<BR>\n" );
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document.write( "3F + 6 - F - 4 = F + 31\r<BR>\n" );
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document.write( "3F - F - F = 31 - 6 + 4\r<BR>\n" );
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document.write( "F  =  29\r<BR>\n" );
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document.write( "Therefore, the 3 consecutive odd integers are: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%2829_31_and_33%29 ALIGN=MIDDLE  ALT=\"highlight_green%2829_31_and_33%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Check<BR>\n" );
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document.write( "3(31) - 33 = 29 + 31\r<BR>\n" );
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document.write( "93 - 33 = 60\r<BR>\n" );
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document.write( "60  =  60 (TRUE)<BR>\n" );
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