document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=393594>Question 393594</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Assuming the general expression for multiples to be 3N, 3(n+1), 3(n+2) etc.\r<BR>\n" );
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document.write( "Find 4 consecutive multiples of 3 such that 5 times the sum of the 1st and 4th is 6 less than 13 times the 3rd.\r<BR>\n" );
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document.write( "I have worked and reworked and can't come up with the right answer.  Can you explain these steps to me?\r<BR>\n" );
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document.write( "Thanks so much! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #279336 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=279336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3n, 3(n+1), 3(n+2), and 3(n+3)\r<BR>\n" );
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document.write( "Then<BR>\n" );
document.write( "5(3n + 3(n+3)) = 13(3(n+2)) - 6<BR>\n" );
document.write( "==> 5(3n + 3n + 9) = 39(n+2) - 6<BR>\n" );
document.write( "==> 5(6n + 9) = 39n + 78 - 6<BR>\n" );
document.write( "==> 30n+ 45 = 39n + 72<BR>\n" );
document.write( "==> -27 = 9n<BR>\n" );
document.write( "==> n = -3\r<BR>\n" );
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document.write( "Then the numbers are -9, -6, -3, 0. </SPAN>\n" );
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