document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1016212>Question 1016212</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One number is 2 times a first number.  A third number is 100 more than the first number.  If the sum of the three numbers is 300&#8203;, find the numbers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #632651 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=632651\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the first number x.  The second number is 2x.  The third number is x+100.  If they add to 300 we write<BR>\n" );
document.write( "x + 2x + x+100 = 300<BR>\n" );
document.write( "4x + 100 = 300<BR>\n" );
document.write( "4x = 200<BR>\n" );
document.write( "x = 50<BR>\n" );
document.write( "2x = 100<BR>\n" );
document.write( "x+100 = 150 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );