+729 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?
+1365 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!:)
+1365 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!:)
