document.write( "Question 133786: Hello,
\n" ); document.write( "I am having a really hard time with these problems. What is the proper way to set up this equation?\r
\n" ); document.write( "\n" ); document.write( "The length of a rectangle is 5 feet more than twice its width. If its area is 52 in², find the dimensions of the length and width.\r
\n" ); document.write( "\n" ); document.write( "Any help would be greatly appreciated.\r
\n" ); document.write( "\n" ); document.write( "Thank you,
\n" ); document.write( "Mrs. Gibson \n" ); document.write( "

Algebra.Com's Answer #97884 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let w=width
\n" ); document.write( "Then length=2w+5\r
\n" ); document.write( "\n" ); document.write( "Area of a rectangle =L*W, so:
\n" ); document.write( "w(2w+5)=52 (w(2w+5)=L*W and we are told that this equals 52 sq in. OK)\r
\n" ); document.write( "\n" ); document.write( "2w^2+5w=52 subtract 52 from each side\r
\n" ); document.write( "\n" ); document.write( "2w^2+5w-52=0 quadratic in standard form\r
\n" ); document.write( "\n" ); document.write( "It's now set up ready to be solved --you can use the quadratic formula\r
\n" ); document.write( "\n" ); document.write( "You will come out with a positive and a negative value for w--neglect the negative value ---lengths and widths in this problem are positive\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps--ptaylor \n" ); document.write( "

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