document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=186840>Question 186840</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 3 more than twice the width.  If the perimeter is to be at least 51 meters, what are the possible values for the width? (If the perimeter is at least 51 meteres, than it is greater than or equal to 51 meters.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #140054 by </B><A HREF=/tutors/aboutme.mpl?userid=uday1435><B>uday1435(57)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=uday1435><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=uday1435><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=140054\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle = 2( l +w)<BR>\n" );
document.write( "Here l = w+3, pluging in that we get p = 2( w+3 + w) = 2(2w+3)\r<BR>\n" );
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document.write( "The given condition is p >= 51<BR>\n" );
document.write( "That is   2(2w + 3) &#8805; 51<BR>\n" );
document.write( "4w + 6 &#8805; 51   adding – 6 on both sides we get<BR>\n" );
document.write( "4w &#8805; 51- 6<BR>\n" );
document.write( "4w &#8805; 45<BR>\n" );
document.write( "Dividing both sides by 4 we get<BR>\n" );
document.write( "w &#8805; 45/4\r<BR>\n" );
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document.write( "w&#8805;  11.25 . So the width must be at least 11.25 meters\r<BR>\n" );
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