Question 212629
Find the dimensions of a rectangle whose length is 40 inches more than its width and whose perimeter is 300 inches.	


Step 1.  Let l = w+40  length of rectangle and let w be the width


Step 2.  Let P = 300 inches be the perimeter.  Perimeter means adding the 4 sides of a rectangle.  So,


{{{P=w+40+w+40+w+w}}}


{{{P=4w+40+40}}}


{{{P=4w+80=300}}}


Step 3.  Subtract 80 to both sides of equation to get 4w by itself



{{{P=4w+80-80=300-80}}}


{{{4w=220}}}


Step 4.  Divide 4 to both sides of equation


{{{4w/4=220/4}}}


{{{w= 55 }}}


Step 5.  w = 55 is the width of the rectangle and length is 95 since


l=55+40=95


Check P=4w+80=4(55)+80=220+80=300  So w = 55 and l = 95 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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