document.write( "Question 385309: The length of a rectangle is 2in. less than 7 times its width. the area of the rectangle is 57 sq. in. What are the dimentions of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #272630 by texttutoring(324)![]() ![]() You can put this solution on YOUR website! Let w = width \n" ); document.write( "Let L = length\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Length is 2 less than 7 times the width:\r \n" ); document.write( "\n" ); document.write( "L = 7w - 2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Formula for area of a rectangle is A = w*L, and we know that A = 57\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Substitute the formula for L into the equation, and solve for w:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A= w*L \n" ); document.write( "57 = w(7w-2) \n" ); document.write( "57 = 7w^2 - 2w \n" ); document.write( "0 = 7w^2 - 2w -57\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Use the quadratic equation (or WolframAlpha.com) to solve.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You should find that w=3\r \n" ); document.write( "\n" ); document.write( "This means that L is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "L=7w-2 \n" ); document.write( "L=7(3)-2 \n" ); document.write( "L=21-2 \n" ); document.write( "L=19\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So the dimensions are w=3, L=19. \n" ); document.write( " |