document.write( "Question 1186358: The length of a wall is 3m more than its witdh.
\n" ); document.write( "If the area of the wall is less than 18m squared,
\n" ); document.write( "What could be its lenght?
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Algebra.Com's Answer #817265 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The problem is far more easily solved informally than with formal algebra.

\n" ); document.write( "6 times 3 is 18, and 3 is 3 less than 6; so the maximum length is 6.

\n" ); document.write( "The minimum length is whatever value keeps the width positive; since the width is 3 less than the length, the minimum length is 3.

\n" ); document.write( "ANSWER: The length can be any x in the interval (3,6)

\n" ); document.write( "Note the length can't be 6, because the area has to be LESS THAN 18 square meters; and it can't be 3, because the width would be 0 and there would be no wall.

\n" ); document.write( "If formal algebra is required....

\n" ); document.write( "x = length
\n" ); document.write( "x-3 = width

\n" ); document.write( "The area (length times width) has to be less than 18:

\n" ); document.write( "\"x%28x-3%29%3C18\"
\n" ); document.write( "\"x%5E2-3x%3C18\"
\n" ); document.write( "\"x%5E2-3x-18%3C0\"
\n" ); document.write( "\"%28x-6%29%28x%2B3%29%3C0\"

\n" ); document.write( "Algebraically, the solution is x between -3 and 6....

\n" ); document.write( "But in the actual problem, since the width has to be positive (x-3>0 --> x>3), the real solution is that x is between 3 and 6.

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