document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=64700>Question 64700</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/093979862x/algebrahomeworkh>An Incremental Development</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> evaluate \r<BR>\n" );
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document.write( "x^2 + xy^3 + xy  if x = - 1/3 and y = -1/2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #45245 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=45245\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2+xy^3+xy<BR>\n" );
document.write( "(-1/3)^2+(-1/3*-1/2^3)+(-1/3*-1/2)<BR>\n" );
document.write( "1/9+(-1/3*-1/8)+(1/6)<BR>\n" );
document.write( "1/9+(1/24)+1/6<BR>\n" );
document.write( "1/9+1/24+1/6 now we need to find a common denominator for these three terms so we can add them <BR>\n" );
document.write( "factors of 9 =3*3<BR>\n" );
document.write( "factors of 6=3*2<BR>\n" );
document.write( "factors of 24=2*2*2*3 combining common factors (using triplicate & duplicate  factors only once) we have 3*3*2*2*2=72<BR>\n" );
document.write( "(8+3+12)/72<BR>\n" );
document.write( "23/72 solution<BR>\n" );
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