A rectangles length is 2 meters greater than its width if the e perimeter is greater than 112 what is the rectangles possible width

Step-by-step explanation:

Perimeter of a rectangle = 2(L +W)        

Given L = W + 2      and the perimeter is greater than 112 meters?

  P rect    <   2(L +W)        L = W + 2

  P rect    <   2(W + 2 +W)

                <   2(2W+2)

  112         <     4W + 4        solve for W

(112 - 4)/4  <   (4W +4 - 4)/4              

108/4       <   (4W + 0)/4

       27     <    W              

      the width has to be greater than 27 meters

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step-by-step explanation:

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set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x