SOLUTION: I have this Geometry/Algebra word problem that I'm not even sure is solvable, I and my entire family have racked our brains over it for hours. Here it goes. a) Define variables

Question 493890: I have this Geometry/Algebra word problem that I'm not even sure is solvable, I and my entire family have racked our brains over it for hours. Here it goes.
a) Define variables
b) Give and equation
c) Give solution
A rectangle is constructed. The length of this rectangle is double the width. Then a new rectangle is made by increasing each side by 3 meters. The perimeter of the new rectangle is 2 meters greater than 4 times the length of the old rectangle. Find the dimensions of the original rectangle.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
A rectangle is constructed. The length of this rectangle is double the width. Then a new rectangle is made by increasing each side by 3 meters. The perimeter of the new rectangle is 2 meters greater than 4 times the length of the old rectangle. Find the dimensions of the original rectangle.

first rectangle:

L = 2x
W = x

second rectangle:

L = 2x + 3
W = x + 3
P = 2(2x+3) + 2(x+3) = 4x+6+2x+6 = 6x+12

L = length
W = width
P = perimeter

P (second rectangle) = 4 * (Length of first rectangle) + 2

this becomes:

6x+12 = 4(2x) + 2 which becomes:
6x+12 = 8x+2
subtract 6x from both sides of the equation to get:
12 = 2x + 2
subtract 2 from both sides of the equation to get:
10 = 2x
divide both sides of the equation by 2 to get:
x = 5

when x = 5:

length of first rectangle becomes 2x = 10
width of first rectangle becomes x = 5

length of second rectangle becomes 2x + 3 = 13
width of second rectangle becomes x + 3 = 8

perimeter of second rectangle becomes 2*13 + 2*8 = 26+16 = 42

4 times length of first rectangle plus 2 becomes 4*10 + 2 = 42

requirements of problem are satisfied.

answer is that the dimensions of the original rectangle are:

length = 10
width = 5

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