Question 1207821
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<pre>
An artist has 50 inches of oak trim to frame a painting. 
The frame is to have a border 3 inches wide surrounding the painting.

(a) If the painting is square, what are its dimensions? 
    What are the dimensions of the frame?

(b) If the painting is rectangular with a length twice its width, what are 
    the dimensions of the painting? What are the dimensions of the frame?
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<pre>
(a)  Imagine this painting surrounded by 3 inches wide frame.

     Let x be size of the painting square, in inches.

     Then from the problem, you have this equation for x

          (x+3) + (x+3) + (x+3) + (x+3) = 50 inches,

     or

          4(x+3) = 50.

     It gives

          x+3 = 50/4 = 12.5 inches,

          x = 12.5 - 3 = 9.5 inches.


     Thus the dimensions of the painting are 9.5 by 9.5 inches.       <<<---===  <U>Answer to (a)</U>.

     Outer dimensions of the frame are 9.5 + 3 + 3 = 15.5 inches.     <<<---===  <U>Answer to (a)</U>.



(b)  Let w be the width of the painting, in inches.

     Then the length of the painting is 2x inches.


     From the problem, you have this equation for x

          (2w+3) + (w+3) + (2w+3) + (w+3) = 50 inches,

     or

          6w+12 = 50.

     It gives

          w = (50-12)/6 = 38/6 = 6{{{1/3}}} inches  (the width),

          2x= 12{{{2/3}}} inches                    (the length).


     Thus the dimensions of the painting are 6{{{1/3}}} inches by 12{{{2/3}}} inches.                           <<<---===  <U>Answer to (b)</U>.

     Outer dimensions of the frame are 6{{{1/3}}} + 3 + 3 = 12{{{1/3}}}  and  12{{{2/3}}} + 3 + 3 = 18{{{2/3}}}  inches.     <<<---===  <U>Answer to (b)</U>.
</pre>

Solved in full.


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Be aware: both the solution method and the answers in the post by @mananth are incorrect.


They are incorrect, because his setup equations are incorrect, in both cases.