document.write( "Question 42285This question is from textbook
\n" ); document.write( ": Here it is copied and pasted exactly out of the book.\r
\n" ); document.write( "\n" ); document.write( "Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?\r
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\n" ); document.write( "\n" ); document.write( "Paul answered it as \r
\n" ); document.write( "\n" ); document.write( "400=(30+2x)(20+20)-(30*20)
\n" ); document.write( "400 = 4x^2 + 100x
\n" ); document.write( "x^2 + 25x - 100 = 0\r
\n" ); document.write( "\n" ); document.write( "Solve that and you get x = 3.5'\r
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\n" ); document.write( "\n" ); document.write( "You replied with this can't be solved, so I am confused now.
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Algebra.Com's Answer #27425 by PaulAllen65270(19)\"\" \"About 
You can put this solution on YOUR website!
It can be solved but I am not complete sure how to do it. What is there is almost correct because the complete area of the garden minus the are of the path would be equal to the remainder of the garden. If you were to draw the garden out with path included you would see that the width would be the same for both path and garden however the length would be shorter for the path then the whole garden by 2x since the path is uniform.
\n" ); document.write( "\"400=%2830%2A20%29-%2820%2830-2x%29%29\"
\n" ); document.write( "\"400=%28600%29-%28600-40x%29\"
\n" ); document.write( "\"400=600-600%2B40x\"
\n" ); document.write( "\"400=40x\"
\n" ); document.write( "\"x=10\" so the width of the path is 10 feet.
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