Question 209740
A rectangle has a perimeter of 14 feet.  If the twice its length is equal to one less than four times its width, find the rectangle's length and width.


Step 1.  Let L be the  length of rectangle and let w be the width


Step 2.  Translate the following into an equation:  If the twice its length is equal to one less than four times its width, find the rectangle's length and width.


{{{2L}}} is twice the length


{{{4w-1}}}  is one less than four time its width.  So now


{{{2L = 4w-1}}}


Divide by 2 to both sides of equation


{{{2L/2=(4w-1)/2}}}


{{{L=2w-(1/2)}}}


Step 3.  Let P = 14  ft be the perimeter.  Perimeter means adding the 4 sides of a rectangle.  So,


{{{P=L+L+w+w=2L+2w}}}


But {{{2L = 4w-1}}}


{{{P=14=4w-1+2w=6w-1}}}


{{{P=6w-1=14}}}


Step 4.  Add 1 to both sides of equation to get 62 by itself



{{{P=6w-1+1=14+1}}}


{{{6w=15}}}


Step 5.  Divide 5 to both sides of equation


{{{6w/6=15/6}}}


{{{w= 5/2 }}}


Step 5.  w = 5/2 is the width of the rectangle and length is 9/2  since



{{{L=2w-(1/2)=2(5/2)-(1/2)}}}


{{{L=9/2}}}


Check P=6w-1= 6*(5/2)-1= 14 So w = 5/2 ft and L=9/2 is the solution.


Hope the above steps were helpful.  Good luck in your homework and studies!  


Respectfully
Dr J


Hope you understood and followed the steps. Good luck in your studies. Dr J 


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