Solve:
This system of equations has one solution. What is z in this problem?
\(\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 3\)
Hi Guest!
\(xy=x+y \\ xz=2(x+z) \\ yz=3(y+z)\)
\(x=\frac{y}{y-1}, \text{ From the first equation}\)
Now, substituting in the second:
\(\frac{y}{y-1}z=2(\frac{y}{y-1}+z)\)
Then, solving for z:
\(yz=2y+2(y-1)z \\ \implies z(y-2(y-1))=2y \\ z=\frac{2y}{2-y}\)
Now, substituting this in the third equation:
\(y(\frac{2y}{2-y})=3(y+\frac{2y}{2-y}) \\ \implies 2y^2=3(y)(2-y)+3(2y) \\ \implies 2y^2=6y-3y^2+6y \\ \implies 5y^2-12y=0 \implies y(5y-12)=0 \implies y_1=0 \text{ or } y_2=\frac{12}{5}\)
Case 1: \(y=0 \implies z=0,x=0\) (Rejected).
Case 2: \(y=\frac{12}{5} \implies z=-12 \implies x=\frac{12}{7}\)
We reject case 1, as it will not satisfy the initial equations given. That is, the denominators will be 0.
Therefore, \((x,y,z):=(\frac{12}{7},\frac{12}{5},-12)\)
Thus, \(z=-12\)
I hope this helps :)!
