document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=15637>Question 15637</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find two consecutive intergers such that the sum of the first and 3 times the second is 55 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #7764 by </B><A HREF=/tutors/aboutme.mpl?userid=pwac><B>pwac(253)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=pwac><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=7764\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let first integer be = x<BR>\n" );
document.write( "so second integer is is x+1<BR>\n" );
document.write( "We now have to develop an equation that matches the question<BR>\n" );
document.write( "So x+3times(x+1)=55 which looks like this:<BR>\n" );
document.write( "x+3(x+1)=55 remove bracket by multiplying by the 3<BR>\n" );
document.write( "x+3x+3=55 which is 4x+3=55 subtract 3 from both sides<BR>\n" );
document.write( "4x=52 divide both sides by 4<BR>\n" );
document.write( "x=13<BR>\n" );
document.write( "x+1=14\r<BR>\n" );
document.write( "\n" );
document.write( "Pete </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );