document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=85497>Question 85497</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One number is 8 less than another. If 10 times the smaller number minus 3 times the larger number is 39, find the two numbers.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #61718 by </B><A HREF=/tutors/aboutme.mpl?userid=rajagopalan><B>rajagopalan(174)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rajagopalan><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=rajagopalan><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=61718\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let first  number = x<BR>\n" );
document.write( "small number = x-8  given<BR>\n" );
document.write( "then 10(x-8)-3x=39 ( 10times smaller minus 3 times bigger)<BR>\n" );
document.write( "10(x-8)-3x=39 <BR>\n" );
document.write( "10x-80-3x=39<BR>\n" );
document.write( "10x-3x=39+80<BR>\n" );
document.write( "7x=119<BR>\n" );
document.write( "x=119/7=17<BR>\n" );
document.write( "so first number = 17<BR>\n" );
document.write( "other number =17-8=9<BR>\n" );
document.write( "Verifucation:<BR>\n" );
document.write( "10times smaller minus 3 times bigger=(10x9)-(3x17)=90-51=39 Ok.<BR>\n" );
document.write( "Answer 17 and 9.<BR>\n" );
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