document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=847878>Question 847878</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find three consecutive positive integers such that that product of the first and third, minus the second, is 1 more than 6 times the third. <BR>\n" );
document.write( "The smallest integer is?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #510725 by </B><A HREF=/tutors/aboutme.mpl?userid=swincher4391><B>swincher4391(1107)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=swincher4391><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=swincher4391><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=510725\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the consecutive integers be known as: n, n+1, and n+2.\r<BR>\n" );
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document.write( "We know that n(n+2) - (n+1) = 1 + 6*(n+2)\r<BR>\n" );
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document.write( "n^2 + 2n - n - 1 = 1 + 6n + 12\r<BR>\n" );
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document.write( "n^2 + n -1 = 6n + 13\r<BR>\n" );
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document.write( "n^2 -5n -14 = 0\r<BR>\n" );
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document.write( "(n-7)(n+2) = 0\r<BR>\n" );
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document.write( "n = 7\r<BR>\n" );
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document.write( "Then the numbers are 7, 8 ,9... 7 being the smallest. </SPAN>\n" );
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