document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=121046>Question 121046</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A number is doubled, and then seven is added. When the result is divided by forty, the quotient is two and the remainder is thirty-three. Find the original number. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #88785 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=88785\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A number is doubled, and then seven is added. When the result is divided by forty, the quotient is two and the remainder is thirty-three. Find the original number.<BR>\n" );
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document.write( "Let the original number be \"x\".<BR>\n" );
document.write( "EQUATION:<BR>\n" );
document.write( "(2x + 7)/40 = 2 +33/40<BR>\n" );
document.write( "2x+7 = 80 + 33<BR>\n" );
document.write( "2x = 106<BR>\n" );
document.write( "x = 103<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );