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\[\large{\begin{cases} x+y+2\sqrt{xy}=64 \\ xy\sqrt{x}+xy\sqrt{y}=32 \end{cases}}\]

Positive real numbers \(x\) and \(y\) satisfy the system of equations above. Find \(x+y\).

 Jan 20, 2021
 #1
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Note that we can  factor the top expression as

 

(sqrt (x) + sqrt (y) ) ^2  = 64              take the  positive root

 

(sqrt (x)  +  sqrt (y) = 8

 

And the  secomd equation cam be factored as

 

xy ( sqrt (x) + sqrt (y) )   =  32

 

xy ( 8)  =  32

 

xy =   32/8   =   4

 

So   using the  first equarion 

 

x + y   +  2sqrt (xy)  =   64

 

x + y +  2sqrt (4)  =  64

 

x +   y  + 2*2  =  64

 

x +  y  +  4  = 64

 

x + y =   64   - 4   =  60

 

 

cool cool cool

 Jan 20, 2021
edited by CPhill  Jan 20, 2021

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