document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=206478>Question 206478</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the value of y for the given value of x by first substituting the value of x into the equation and solving for y.<BR>\n" );
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document.write( "xy+x=-3; x=3<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 #156023 by </B><A HREF=/tutors/aboutme.mpl?userid=rfer><B>rfer(16322)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rfer><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=156023\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> xy+x=-3<BR>\n" );
document.write( "3y+3=-3<BR>\n" );
document.write( "subtract 3 from both sides<BR>\n" );
document.write( "3y+3-3=-3-3<BR>\n" );
document.write( "3y=-6<BR>\n" );
document.write( "divide both sides by 3<BR>\n" );
document.write( "3y/3=-6/3<BR>\n" );
document.write( "y=-2 </SPAN>\n" );
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