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Here's a problem I have been trying to figure out. I can't seem to find out what the answer is though...

 

I have a picture with dimensions \(x\) and \(y\) (in inches), such that \(x\) and \(y\) are both integers greater than one. I would like to place this picture in an elongated frame of dimensions \(2x+3\) and \(y+2\). If I measured the area of the frame to be \(34\) square inches, what is the area of the picture in square inches? (Note that by "the area of the frame," we mean the shaded region shown below).

 

 

I think using the complete the square method works. Does it?

 Apr 21, 2019
edited by cantthinkofausername  Apr 21, 2019
edited by cantthinkofausername  Apr 22, 2019
 #1
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Hmmmmm .....     y - 2    would make the frame smaller than the picture.....are you sure it is y - 2?

 Apr 21, 2019
 #2
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Oh yeah, ill change it.

cantthinkofausername  Apr 22, 2019
 #3
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There is a RANGE of solutions for this....

 

Area of ENTIRE frame - area of picture = 34

(2x+3)(y+2)                   -   xy                  = 34

xy +4x +3y = 28   

See diagram of this equation graphed on desmos.....pick any (positive) x,y on the graph....it will solve this equation...

   I picked a couple of possibilities    x,y       5,1    2,4      2.844, 2.844     etc......

 

 Apr 22, 2019

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