pre-algebra word problem. Help Please

Question

Without actually solving the problem, choose the correct solution by deciding which choice satisfies the given conditions. The length of a rectangle is 2 feet longer than the width. The perimeter is 32 feet. Find the dimensions of the rectangle.

Answer 1

The mathematical model for this problem is the pair of equations:

 

2(L + W) = 32

L = W + 2

 

 

Answer 2

If we let the width of the rectangle = x, then the length of the rectangle = x + 2

 

The formula for the perimeter P of a rectangle is P = 2L + 2W, where L= length and W = width

 

If we substitute into the above formula our x-values for the length and the width, along with the given perimeter of this rectangle, we have:

 

32 = 2(x + 2) + 2x

 

32 = 2x + 4 + 2x

 

32 = 4x + 4

 

4x + 4 = 32

Answer 3

You do not show what choices you have, but anyway, if the perimeter 2(L+W) = 32, then L+W must be 16.

 

What two positive integers differ by two and sum to 16?