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

 Jun 22, 2024
 #1
avatar+129725 
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xy  = x + y   →  xy - x  = y →  x ( y - 1)  = y →  x = y / (y  - 1)

xz =  x +  z

yz =  y + z   → yz - z = y → z (y - 1) = y  → z = y/(y - 1)

 

x = z

 

So

 

xz = x + z

z*z  = z + z

z^2  = 2z

z^2 - 2z = 0 

z ( z  - 2)  = 0 

 

z = 0    reject

 

z = 2 

 

cool cool cool

 Jun 22, 2024

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