document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=154933>Question 154933</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you figure this out when y is isolated on both equations:<BR>\n" );
document.write( "y=3x+1<BR>\n" );
document.write( "y=6x-1\r<BR>\n" );
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document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #114065 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=114065\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you substitute equals for equals\r<BR>\n" );
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document.write( "3x+1=6x-1 __ subtracting 3x __ 1=3x-1 __ adding 1 __ 2=3x __ dividing by 3 __ 2/3=x\r<BR>\n" );
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document.write( "substituting for x __ y=3(2/3)+1 __ y=3 </SPAN>\n" );
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