document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=994145>Question 994145</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the number such that when 3/4 of it is added to 3whole no 1/2 the sum is the such a when 2/3 of it is subtracted from 6whole no 1/2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #613355 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=613355\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the number such that when 3/4 of it is added to 3whole no 1/2 the sum is the such a when 2/3 of it is subtracted from 6whole no 1/2 <BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "(3/4)x + 3 1/2 = 6 1/2 - (2/3)x<BR>\n" );
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document.write( "(3/4)x + (7/2) = (13/2) - (2/3)x<BR>\n" );
document.write( "----<BR>\n" );
document.write( "[(3/4)+(2/3)]x = (13/2)-(7/2)<BR>\n" );
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document.write( "[(9-8)/12]x = 3<BR>\n" );
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document.write( "x = 36<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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