SOLUTION: The square poster has sides measuring 2 feet less than the sides of a square sign. If the distance between their area is 52 square feet find the lengths of the sides of the poster

Algebra ->  Surface-area -> SOLUTION: The square poster has sides measuring 2 feet less than the sides of a square sign. If the distance between their area is 52 square feet find the lengths of the sides of the poster       Log On


   

Question 1060201: The square poster has sides measuring 2 feet less than the sides of a square sign. If the distance between their area is 52 square feet find the lengths of the sides of the poster and the sign.
Answer by ikleyn(52781) About Me  (Show Source):
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The square poster has sides measuring 2 feet less than the sides of a square sign. If the distance between their area is 52 square feet
find the lengths of the sides of the poster and the sign.
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Answer. 12 cm (poster) and 14 cm (sign). 

x - y = 2,          (1)
x^2 - y^2 = 52.     (2)

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x - y = 2,          (1')
(X+y)*(x-y) = 52.   (2')

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x - y = 2,          (1'')
x + y = 26.         (2'')


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add (1'') and (2'') to get  x = 14.

Then y = 14 - 2 = 12.

Solved.