document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1008684>Question 1008684</A>: <I><SPAN STYLE=\"word-break: break-all;\"> one positive number is 6 less than twice a second number, and the product is 140. Find the two numbers\r<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 #624281 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=624281\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the numbers x and y.  Thus we have<BR>\n" );
document.write( "xy = 140<BR>\n" );
document.write( "and<BR>\n" );
document.write( "x = 2y - 6<BR>\n" );
document.write( "Now substitute into the first equation and get<BR>\n" );
document.write( "(2y - 6)(y) = 140<BR>\n" );
document.write( "2y^2 - 6y - 140 = 0<BR>\n" );
document.write( "y^2 - 3y - 70 = 0<BR>\n" );
document.write( "(y - 10)(y + 7) = 0<BR>\n" );
document.write( "y = 10  or y = 7<BR>\n" );
document.write( "but it needs to be positive so<BR>\n" );
document.write( "y = 10  and then<BR>\n" );
document.write( "x = 14 </SPAN>\n" );
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