document.write( "Question 375738: the length of a retangle is 3 less than 5 times of its width.
\n" ); document.write( "write a simplified algebraic expression for the perimeter of a rectangle.
\n" ); document.write( "if the rectangle width is tripled and its length is doubled the perimeter of new rectangle is 92 cm greater than original perimeter.
\n" ); document.write( "Find the area of the original rectangle. \n" ); document.write( "
Algebra.Com's Answer #267195 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
let width =x
\n" ); document.write( "length = 5x-3
\n" ); document.write( "..
\n" ); document.write( "Perimeter = 2(x+5x-3)
\n" ); document.write( "Perimeter = 2(6x-3)
\n" ); document.write( "=12x-6
\n" ); document.write( "...
\n" ); document.write( "new rectangle
\n" ); document.write( "width = 3x
\n" ); document.write( "length = 2(5x-3)
\n" ); document.write( "perimeter = 2(3x+2(5x-3))
\n" ); document.write( "perimeter = 2(3x+10x-6)
\n" ); document.write( "perimeter = 6x+20x-12
\n" ); document.write( "...
\n" ); document.write( "6x+20x-12=12x-6+92
\n" ); document.write( "6x+20x-12x=-6+92+12
\n" ); document.write( "14x=98
\n" ); document.write( "/14
\n" ); document.write( "x=98/14
\n" ); document.write( "x=7 cm. the width of original rectangle
\n" ); document.write( "length = 5x-3 = 5*7-3 = 32cm
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "
\n" );