document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366416>Question 366416</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do I solve using the substituion method with the following equations. 3b-2a=4 and b/2-2a/3=1?\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 #261108 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=261108\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  3b-2a=4<BR>\n" );
document.write( "3b= 4+2a<BR>\n" );
document.write( "3b=2(2+a)<BR>\n" );
document.write( "b=2(2+a)/3\r<BR>\n" );
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document.write( "....\r<BR>\n" );
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document.write( " and b/2-2a/3=1<BR>\n" );
document.write( "Plug value of b in the baove equation<BR>\n" );
document.write( "2(2+a)/3*2 - 2a/3 =1\r<BR>\n" );
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document.write( "(2+a)/3 -2a/3 =1<BR>\n" );
document.write( "...<BR>\n" );
document.write( "(2+a-2a)/3 =1<BR>\n" );
document.write( "(2-a) /3=1<BR>\n" );
document.write( "2-a=3<BR>\n" );
document.write( "a=2-3<BR>\n" );
document.write( "a=-1\r<BR>\n" );
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document.write( "plug value of a in the equation\r<BR>\n" );
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document.write( "b=2(2+a)/3<BR>\n" );
document.write( "b=2(2-1)/3<BR>\n" );
document.write( "b=2/3<BR>\n" );
document.write( "...<BR>\n" );
document.write( "m.ananth@hotmail.ca </SPAN>\n" );
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