+84 The system of equations
\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 1, \quad \frac{yz}{y + z} = 2
has exactly one solution. What is $z$ in this solution?
+2 We are given the system of equations:
\(\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 1, \quad \frac{yz}{y + z} = 2\)
Let’s start by simplifying the equations one by one.
From the first equation: \(\frac{xy}{x+y}=1\)
This implies:
\(xy=x+y\)
Rearranging:
\(xy-x-y=0\)
Factorizing:
\((x-1)(y-1)=1\)
From the second equation:
\(\frac{xz}{x+z}=1\)
This implies:
\(xz=x+z\)
Rearranging:
\(xz-x-z=0\)
Factorizing:
\((x-1)(z-1)=1\)
From the third equation:
\(\frac{yz}{y+z}=2\)
This implies:
\(yz=2(y+z)\)
Expanding:
\(yz=2y+2z\)
Rearranging:
\(yz-2y-2z=0\)
Factorizing:
\((y-2)(z-2)=4\)
We now have the following system of equations:
\((x-1)(y-1)=1\\ (x-1)(z-1)=1\\ (y-2)(z-2)=4\)
From equations 1 and 2, we notice that:
\((x-1)(y-1)=(x-1)(z-1)=1\)
Since both are equal to 1, we can conclude that:
\(y-1=z-1\)
This gives us:
\(y=z\)
Substituting y = z into the third equation:
\((z-2)(z-2)=4\)
Expanding:
\((z-2)^2=4\)
Taking the square root of both sides:
\(z-2=\pm2\)
Therefore,
\(z=4\) or \(z=0\)
If \(z=4\), then \(y=4\). Substituting \(y=4\) into equation 1:
\((x-1)(4-1)=1\\ (x-1)(3)=1\\ x-1=\frac{1}{3}\\ x=\frac{4}{3}\)
So, \(x=\frac{4}{3}\), \(y=z\), and \(z=4\) is a solution.
If \(z=0\), then \(y=0\). Substituting \(y=0\) into equation 1:
\((x-1)(0-1)=1\\ (x-1)(-1)=1\\ x-1=-1\\ x=0\)
So, \(x=0\), \(y=0\), and \(z=0\) is a possible solution. However, substituting into the second equation:
\( \frac{0 \times 0}{0+0}\nexists\), which is not valid. So, \(z=0\) is not a valid solution.
The only valid solution is \(x=\frac{4}{3}\), \(y=4\), and \(z=4\). Therefore, the value of z is: \(z=4\).
