<PRE><B>Question 1186358</B>
&lt;br&gt;
The problem is far more easily solved informally than with formal algebra.&lt;br&gt;
6 times 3 is 18, and 3 is 3 less than 6; so the maximum length is 6.&lt;br&gt;
The minimum length is whatever value keeps the width positive; since the width is 3 less than the length, the minimum length is 3.&lt;br&gt;
ANSWER: The length can be any x in the interval (3,6)&lt;br&gt;
Note the length can&#39;t be 6, because the area has to be LESS THAN 18 square meters; and it can&#39;t be 3, because the width would be 0 and there would be no wall.&lt;br&gt;
If formal algebra is required....&lt;br&gt;
x = length
x-3 = width&lt;br&gt;
The area (length times width) has to be less than 18:&lt;br&gt;
{{{x(x-3)&lt;18}}}
{{{x^2-3x&lt;18}}}
{{{x^2-3x-18&lt;0}}}
{{{(x-6)(x+3)&lt;0}}}&lt;br&gt;
Algebraically, the solution is x between -3 and 6....&lt;br&gt;
But in the actual problem, since the width has to be positive (x-3&gt;0 --&gt; x&gt;3), the real solution is that x is between 3 and 6.&lt;br&gt;
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