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The system of equations
{xy}/{x + y} = 1,{xz}/{x + z} = 1, {yz}/{y + z} = 3
has exactly one solution. What is z in this solution?

 Apr 2, 2024
 #1
avatar+129690 
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(xy) / (x + y)  = 1

(xz) /(x + z) =1

(yz) ( y + z)  = 3

 

xy = x + y  

xz = x + z   →   xz - x = z  →  x ( z - 1)  = z →  x =  z / (z -1)

yz =  3(y + z)  →   yz = 3y + 3z →  yz - 3y = 3z →  y ( z - 3)  =  3z →  y = 3z / ( z -3)

 

 

xy  = x + y

 

[ z / ( z -1) ] [3z / ( z -3) ] = z / (z - 1) + 3z / ( z -3)

 

[ 3z^2 ] / [ (z -1) ( z - 3) ] =  z (z -3) + 3z ( z -1) / [ (z -3) (z -1) ]

 

3z^2  = z ( z -3) +3z ( z -1)

 

3z^2  = z^2 - 3z + 3z^2 - 3z

 

z^2 - 6z  =  0

 

z ( z - 6 ]  = 0

 

z = 0   (reject)

 

z =  6

 

y = 6

 

x  = 6/5

 

cool cool cool

 Apr 4, 2024

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