document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4962>Question 4962</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How would I solve the compound inequality and write the solution set in interval notation?\r<BR>\n" );
document.write( "\n" );
document.write( "-1 < 3 - 2x < 11 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2463 by </B><A HREF=/tutors/aboutme.mpl?userid=Abbey><B>Abbey(339)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Abbey><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=2463\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Split this into two parts:\r<BR>\n" );
document.write( "\n" );
document.write( "-1<3-2x  and 3-2x<11<BR>\n" );
document.write( "-1<3-2x:<BR>\n" );
document.write( "Subtract 2x from both sides of the equation:<BR>\n" );
document.write( "-4<-2x<BR>\n" );
document.write( "Divide by -2 (you have to switch the inequality sign when dividing by a negative number)<BR>\n" );
document.write( "2>x\r<BR>\n" );
document.write( "\n" );
document.write( "Second part:<BR>\n" );
document.write( "3-2x<11<BR>\n" );
document.write( "subtract 3 from both sides:<BR>\n" );
document.write( "-2x<8<BR>\n" );
document.write( "Divide by -2 (don't forget to switch the sign)<BR>\n" );
document.write( "x>-4\r<BR>\n" );
document.write( "\n" );
document.write( "Now we see that in order to have this be a true statement the value for x has to be greater than negative 4, but less than 2, so the solution set is:<BR>\n" );
document.write( "(-4,2)\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );