document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=87085>Question 87085</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0536818932/algebrahomeworkh>College Algebra Math 107</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I am stuck on this problem.  Please help.  The perimeter of a rectangle is 34 feet and its area is 60 square feet.  Find the length and width of the rectangle.\r<BR>\n" );
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document.write( "Here is what I did:<BR>\n" );
document.write( "I used the P = 2l + 2w formula<BR>\n" );
document.write( "34 = 2l + 2w.<BR>\n" );
document.write( "At this point I'm stuck because I'm not sure how to incorporate the area (60 sq. ft.) into the problem. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #63052 by </B><A HREF=/tutors/aboutme.mpl?userid=tutorcecilia><B>tutorcecilia(2152)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tutorcecilia><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=tutorcecilia><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=63052\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the formula for the Area:<BR>\n" );
document.write( "Area = (length)(width)<BR>\n" );
document.write( "Area = 60 [plug-in the values that are given]<BR>\n" );
document.write( "Sovling for either L or W will be too confusing, so...<BR>\n" );
document.write( ".<BR>\n" );
document.write( "Use the formula for Perimeter:<BR>\n" );
document.write( "P=2(length + width)<BR>\n" );
document.write( "P=34 [plug-in the values that are given]<BR>\n" );
document.write( "34=2(length + width)<BR>\n" );
document.write( "34/2=2(length + width)/2 [solve for either the Width or the Length]<BR>\n" );
document.write( "17=length + width<BR>\n" );
document.write( "17-width=length+width-width<BR>\n" );
document.write( "17-width=length<BR>\n" );
document.write( "17-W=L<BR>\n" );
document.write( ".<BR>\n" );
document.write( "Area=(L)(W)<BR>\n" );
document.write( "60=(17-W)(W) [plug-in the values and simplify]<BR>\n" );
document.write( "60=17W-w^2 [set the equation equal to zero by adding/subtraction from both sides<BR>\n" );
document.write( "60-17W+w^2=-17W+17W-W^2+W^2<BR>\n" );
document.write( "60-17W+W^2=0 [re-arrange the terms]<BR>\n" );
document.write( "w^2-17W+60=0 [factor]<BR>\n" );
document.write( "(W-12)(W-5)=0 [set each factor equal to zero and solve for the W-term]<BR>\n" );
document.write( "W-12=0<BR>\n" );
document.write( "W=12<BR>\n" );
document.write( "or <BR>\n" );
document.write( "W-5=0<BR>\n" );
document.write( "W=5<BR>\n" );
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document.write( "Since the width is the smallest of the 5 and 12, let 5=width and let 12 = length.\r<BR>\n" );
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document.write( "Check by plugging everything back into both equations.<BR>\n" );
document.write( "A=LW<BR>\n" );
document.write( "60=(12)(5)<BR>\n" );
document.write( "60=60 [checks out]<BR>\n" );
document.write( "and<BR>\n" );
document.write( "P=2(length + width)<BR>\n" );
document.write( "34=2(12+5)<BR>\n" );
document.write( "34=2(17)<BR>\n" );
document.write( "34=34 [also checks out] </SPAN>\n" );
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