document.write( "Question 1102565: the length of a rectangle is 4 centimeters more than the width. If both the width and the length were tripled, the area would be 108 square centimeters. Find the dimensions of the orginal rectangle
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Algebra.Com's Answer #717310 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "When the length and width are both tripled (multiplied by 3), the area is increased by a factor of 3*3=9. So the original area was 108/9=12.

\n" ); document.write( "Inspection shows that two numbers that differ by 4 and have a product of 12 are 2 and 6; those are the dimensions of the original rectangle.

\n" ); document.write( "If you need the formal algebra, it might look like this....

\n" ); document.write( "let x = original width
\n" ); document.write( "then x+4=original length
\n" ); document.write( "3x = new width
\n" ); document.write( "3(x+4) = 3x+12 = new length

\n" ); document.write( "The new area (length times width) is 108:
\n" ); document.write( "\"%283x%29%283x%2B12%29+=+108\"
\n" ); document.write( "\"9x%5E2%2B36x+=+108\"
\n" ); document.write( "\"9x%5E2%2B36x-108+=+0\"
\n" ); document.write( "\"x%5E2%2B4x-12+=+0\"
\n" ); document.write( "\"%28x%2B6%29%28x-2%29+=+0\"
\n" ); document.write( "x = -6 or x = 2;
\n" ); document.write( "reject the negative solution since it is the width\n" ); document.write( "The original width was x=2; the original length was x+4=6.

\n" ); document.write( "Learning how to solve the problem using formal algebra is good training, for when the problems get too complicated to solve informally by logical reasoning.

\n" ); document.write( "But you should also be able to use your powers of logical reasoning to solve simpler problems like this, without formal mathematical methods.
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