document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=60511>Question 60511</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0072549017/algebrahomeworkh>Begining Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> From Beginning Algebra Huchison online edition from UOP Chapter 2 section 3 problem 54<BR>\n" );
document.write( "6x-1/5 -2x/3 =3\r<BR>\n" );
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document.write( "I know I am supposed to get rid of the fractions by multiplying LCM times terms, but does that include the =3? With or without I am so very lost. Your great explanations would be greatly appreciated!<BR>\n" );
document.write( "Thanks<BR>\n" );
document.write( "Kelley </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #41554 by </B><A HREF=/tutors/aboutme.mpl?userid=kayeith><B>kayeith(17)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kayeith><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=41554\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 6x - 1/5 - 2x/3 = 3<BR>\n" );
document.write( "(6x - 1/5 - 2x/3 = 3)15<BR>\n" );
document.write( "90x - 3 - 10x = 15<BR>\n" );
document.write( "80 x = 15 + 3<BR>\n" );
document.write( "80x = 18<BR>\n" );
document.write( "x = 18/80<BR>\n" );
document.write( "x = 9 / 40 </SPAN>\n" );
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