document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=288390>Question 288390</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the length and width of a rectangle if the perimeter is 24m and the area is 35m2? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #209037 by </B><A HREF=/tutors/aboutme.mpl?userid=jaydducote><B>jaydducote(11)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jaydducote><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=jaydducote><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=209037\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Area=length times width, or A=LW<BR>\n" );
document.write( "Perimeter=2W+2L<BR>\n" );
document.write( "We know the perimeter is 24m, so 24=2W+2L<BR>\n" );
document.write( "We know the area is 35 square meters, so 35=LW<BR>\n" );
document.write( "Now you need to do substitution to solve.  You have to solve for either L or W in terms of the other variable.  I would solve the first one for W:<BR>\n" );
document.write( "24=2W+2L<BR>\n" );
document.write( "subtract 2L from both sides<BR>\n" );
document.write( "24-2L=2W<BR>\n" );
document.write( "now divide both sides by 2<BR>\n" );
document.write( "12-L=W<BR>\n" );
document.write( "so we know that W=12-L.  Substitute that into the other equation:<BR>\n" );
document.write( "35=LW<BR>\n" );
document.write( "35=L(12-L)<BR>\n" );
document.write( "distribute the L<BR>\n" );
document.write( "35=12L-L^2<BR>\n" );
document.write( "This now looks tricky, there is a square term, so we will need to factor<BR>\n" );
document.write( "Move everything to the left side<BR>\n" );
document.write( "L^2-12L+35=0<BR>\n" );
document.write( "Now, think about which two numbers add to twelve and multiply to get 35.  Think of the factors of 35.  Only 7 times 5 works, which happen to add to 12!<BR>\n" );
document.write( "Since we have a -12L but a +35, and we know we are adding to get to 12, not subtracting, we should conclude that both the 7 and 5 must be negative.  Factored, we get this:<BR>\n" );
document.write( "(L-5)(L-7)=0<BR>\n" );
document.write( "This means that our two \"zeros\" or solutions are the numbers that make this statement true.  If L is 5 or 7, the entire value is 0, so those are our answers.  W=5, L=7. </SPAN>\n" );
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