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## Problema 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 four times the length of the old rectangle. Find the dimensions of the original rectangle

## Answer provided by our tutors

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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