document.write( "Question 519099: The length of a rectangle is 3 yds longer than its width.
\n" ); document.write( "If the perimeter of the rectangle is 42 yds , find its area. \n" ); document.write( "
Algebra.Com's Answer #345382 by mamiya(56)\"\" \"About 
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let's p be the perimeter, a the area, l the length and w the width .\r
\n" ); document.write( "\n" ); document.write( " since we are talking about a rectangle, p=2(l+w) and a=lw\r
\n" ); document.write( "\n" ); document.write( " l is 3 yds longer than w, means l=w+3
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\n" ); document.write( " p=2(l+w)
\n" ); document.write( " =2(w+3+w)
\n" ); document.write( " =4w +6
\n" ); document.write( "from this we have w= (p-6)/4
\n" ); document.write( " At this level, there are two possibilities.
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\n" ); document.write( " the first one, find the value of l and w and then a
\n" ); document.write( " w=(p-6)/4= (42-6)/4 = 9 yds
\n" ); document.write( " l= w+3= 9+3 =12 yds
\n" ); document.write( "then a= lw= 12*9= 108 yds^2\r
\n" ); document.write( "\n" ); document.write( " the second one, express w as a function of p and and we change the l into w+3 and we put both of them in the formula of the area\r
\n" ); document.write( "\n" ); document.write( " a=lw
\n" ); document.write( " =(w+3)w we know p=4w+6 , so w= (p-6)/4
\n" ); document.write( " = [(p-6)/4 +3] (p-6)/4
\n" ); document.write( " = [ (42-6)/4 +3] (42-6)/4
\n" ); document.write( " = 108 yds^2
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