document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=881102>Question 881102</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sum of three consecutive nos is 333. find the sum of last two numbers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #531948 by </B><A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><B>algebrahouse.com(1659)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=algebrahouse.com><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=algebrahouse.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=531948\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x = first number<BR>\n" );
document.write( "x + 1 = second number<BR>\n" );
document.write( "x + 2 = third number\r<BR>\n" );
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document.write( "x + x + 1 + x + 2 = 333   {sum of the three consecutive numbers is 333}<BR>\n" );
document.write( "3x + 3 = 333   {combined like terms}<BR>\n" );
document.write( "3x = 330 {subtracted 3 from each side}<BR>\n" );
document.write( "x = 110 {divided each side by 3}<BR>\n" );
document.write( "x + 1 = 111   {substituted 110, in for x, into x + 1}<BR>\n" );
document.write( "x + 2 = 112 {substituted 110, in for x, into x + 2}\r<BR>\n" );
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document.write( "sum of last two numbers <BR>\n" );
document.write( "= 111 + 112<BR>\n" );
document.write( "= 223<BR>\n" );
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