document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=660114>Question 660114</A>: <I><SPAN STYLE=\"word-break: break-all;\"> After a certain number is diminished by 8, the amount remaining is multiplied by 8; the result is the same if the number were diminished by 6 and the amount remaining multiplied by 6. What is the number? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #411121 by </B><A HREF=/tutors/aboutme.mpl?userid=kevwill><B>kevwill(135)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kevwill><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=kevwill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=411121\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the number we're looking for.  Then, from the problem statement:<BR>\n" );
document.write( "8*(x-8) = 6*(x-6)<BR>\n" );
document.write( "Expanding:<BR>\n" );
document.write( "8x - 64 = 6x - 36<BR>\n" );
document.write( "Subtracting 6x - 64 from both sides:<BR>\n" );
document.write( "8x - 64 - (6x - 64) = 6x - 36 - (6x - 64)<BR>\n" );
document.write( "8x - 64 - 6x + 64 = 6x - 36 - 6x + 64<BR>\n" );
document.write( "2x = 18<BR>\n" );
document.write( "x = 14<BR>\n" );
document.write( "Checking:<BR>\n" );
document.write( "8*(14 - 8) = 8*6 = 48<BR>\n" );
document.write( "6*(14 - 6) = 6*8 = 48\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );