The width (w) of a rectangle is 2 more than one-third its length (l). If the perimeter is 124 cm, identify the equation for the problem.

Question

write the equation 

Answer 1

Let w=width and L=length

 

w = 2+ 1/3 L         and P = 2w + 2L = 124

 

substitute what they gave you for w into the second equation

 

2w + 2L = 124

2(2+ 1/3 L) + 2L = 124

4 + 2/3L + 2L = 124       let's get rid of that pesky fraction by multiplying everything by 3

12 + 2L + 6L = 372

12 + 8L = 372

-12          -12

       8L = 360

      /8       /8

       L  =  45

 

now put that value for L in one of the equations and solve for w

 

w = 2+ 1/3 L

w = 2 + 1/3 ( 45)

w = 2 + 15

w = 17