document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=397751>Question 397751</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the following linear equation:<BR>\n" );
document.write( "(x+1)/(2x-1)=4/3<BR>\n" );
document.write( "Thanks in advance. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #281809 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=281809\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solve the following linear equation:<BR>\n" );
document.write( "(x+1)/(2x-1)=4/3<BR>\n" );
document.write( "Thanks in advance.\r<BR>\n" );
document.write( "\n" );
document.write( "When you have only two terms like the given expression, the best way to start is to cross multiply, that is, use the old rule,\"product of the means=the product of the extremes\"\r<BR>\n" );
document.write( "\n" );
document.write( "x+1)/(2x-1)=4/3<BR>\n" );
document.write( "3x+3=8x-4<BR>\n" );
document.write( "5x=7<BR>\n" );
document.write( "x=7/5\r<BR>\n" );
document.write( "\n" );
document.write( "The conventional way  to start is to eliminate fractions by multiplying each term by the LCD<BR>\n" );
document.write( "The LCD in this equation is 3(2x-1). As you can see, it requires a little more work.  However, when you have several or more terms in the equation, this is the method to use. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );