document.write( "Question 86695: Solve algebraically using only one variable. The length of a rectangle is two more than twice its width. If the area of the rectangle is 84, find the length and width. \n" ); document.write( "
Algebra.Com's Answer #62716 by Flake(45)\"\" \"About 
You can put this solution on YOUR website!
--Given#1: \"length (L) is two more than twice its width (W)\"
\n" ); document.write( "What this means is---> L = 2+2W
\n" ); document.write( "--Given#2: Area of the rectangle is 84
\n" ); document.write( "What this means is --> Area of rectangle = LW = 84\r
\n" ); document.write( "\n" ); document.write( "--Solution:
\n" ); document.write( "--Step#1: Start with Area: LW=84
\n" ); document.write( "--Step#2: Replace L with Given#1 ----> (2+2W)W = 84
\n" ); document.write( "--Step#3: Expand (2+2W)W = 84
\n" ); document.write( "2W+2W^2 = 84
\n" ); document.write( "2(W+W^2) = 84
\n" ); document.write( "W+W^2 = 42
\n" ); document.write( "W^2 + W -42 = 0
\n" ); document.write( "(W+7)(W-6) = 0
\n" ); document.write( "W= \"-7\" or 6
\n" ); document.write( "Since length of rectangle can't be \"negative\", then W must be 6\r
\n" ); document.write( "\n" ); document.write( "--Step#4: Substitude W=6 to Given#2
\n" ); document.write( "LW=84
\n" ); document.write( "L(6)=84
\n" ); document.write( "L=84/6
\n" ); document.write( "L=14\r
\n" ); document.write( "\n" ); document.write( "--Check#1: L = 2+2W
\n" ); document.write( "14 = 2+2(6)---Yes
\n" ); document.write( "--Check#2: LW=84
\n" ); document.write( "14(6)=84---Yes
\n" ); document.write( " \n" ); document.write( "
\n" );