document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=90215>Question 90215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sum of the third and twice the first of three consecutive odd integers is 49 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #65508 by </B><A HREF=/tutors/aboutme.mpl?userid=rossiv53><B>rossiv53(27)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rossiv53><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=65508\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = 1st number<BR>\n" );
document.write( "(x+2) = 2nd number<BR>\n" );
document.write( "(x+4) = 3rd number\r<BR>\n" );
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document.write( "Since the question asks for the sum of twice the first number and the third number, we will add those two values.\r<BR>\n" );
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document.write( "(x+4)+2x=49<BR>\n" );
document.write( "3x+4=49<BR>\n" );
document.write( "3x+4-4=49-4<BR>\n" );
document.write( "3x=45<BR>\n" );
document.write( "x=15\r<BR>\n" );
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document.write( "So 2x = 30.  30 + your 3rd number(19) equals 49.\r<BR>\n" );
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document.write( "Good luck! </SPAN>\n" );
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