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# System

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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

#1
+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
+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!:)

NotThatSmart Jun 7, 2024