document.write( "Question 144909: I need some help with this word problem. Any help would be greatly appreciated. Thanks.\r
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\n" ); document.write( "\n" ); document.write( "A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo?
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Algebra.Com's Answer #105601 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!
Sure! Let's start with the unknown ---- width = \"x\".
\n" ); document.write( "And we know the length is 3 inches longer, so length= \"x+3\" .
\n" ); document.write( "Now you attached 2-inch border around it, meaning you extended the orig. width 2inches on the left side and 2 inches on the right side right? So the new width dimension will be \"x+2+2\"= \"x+4\". Likewise the length is extended 2 inches on the top and 2 inches on the bottom, so it's new dimension will be \"x+3+2+2\" = \"x+7\".
\n" ); document.write( "Here we gonna use the area given with the photo and the border plus the new dimensions right? So going back in finding the area of rectangle,
\n" ); document.write( "A= (x+4)(x+7) ------------ eqn 1
\n" ); document.write( "108= x^2 + 11x + 28
\n" ); document.write( "x^2 + 11x -80 = 0
\n" ); document.write( "so (x+16)(x-5) right?
\n" ); document.write( "It has 2 values, x= -16 which we can't use being negative
\n" ); document.write( "x= 5... perfect! This the one to use.
\n" ); document.write( "Therefore going back to orig. dimension we can get the size of the photo, where the width = x= 5 inches. Also the length = x+3=5+3= 8inches.
\n" ); document.write( "In doubt? Go back eqn 1
\n" ); document.write( "A= (5+4)(5+7)
\n" ); document.write( "108in^2 = 9*12
\n" ); document.write( "108in^2 = 108in^2
\n" ); document.write( "thank you,
\n" ); document.write( "Jojo\r
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