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The system of equations

\frac{xy}{x + y} = 2, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 2

has exactly one solution. What is z in this solution?

 Jun 7, 2024

Best Answer 

 #1
avatar+1280 
+1

First up, let's isolate x for \(\frac{xy}{x+y}=2:\\ \)

We get \(x=\frac{2y}{y-2}\)

Subbing this in, we get 

\(\frac{\frac{2y}{y-2}z}{\frac{2y}{y-2}+z}=2\\ \frac{yz}{y+z}=2\)

\(\frac{2yz}{2y+z\left(y-2\right)}=2\\ \frac{yz}{y+z}=2\)

 

Now, we isolate y. 

\(\frac{2yz}{2y+z\left(y-2\right)}=2 \\ y=z\)

 

Subsituting in y=z, we get

\(\frac{zz}{z+z}=2\)

\(z=4\)

 

We get z=4 as our answer. 

 

Thanks!:)

 Jun 7, 2024
 #1
avatar+1280 
+1
Best Answer

First up, let's isolate x for \(\frac{xy}{x+y}=2:\\ \)

We get \(x=\frac{2y}{y-2}\)

Subbing this in, we get 

\(\frac{\frac{2y}{y-2}z}{\frac{2y}{y-2}+z}=2\\ \frac{yz}{y+z}=2\)

\(\frac{2yz}{2y+z\left(y-2\right)}=2\\ \frac{yz}{y+z}=2\)

 

Now, we isolate y. 

\(\frac{2yz}{2y+z\left(y-2\right)}=2 \\ y=z\)

 

Subsituting in y=z, we get

\(\frac{zz}{z+z}=2\)

\(z=4\)

 

We get z=4 as our answer. 

 

Thanks!:)

NotThatSmart Jun 7, 2024

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