+35 The system of equations
has exactly one solution. What is \(z\) in this solution?
+26393 The system of equations
\(\dfrac{xy}{x+y}=1,\ \dfrac{xz}{x+z}=2,\ \dfrac{yz}{y+z}=3 \)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{xy}{x+y}} &=& \mathbf{1} \\\\ \dfrac{x+y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{xz}{x+z}} &=& \mathbf{2} \\\\ \dfrac{x+z} {xz}&=& \dfrac{1}{2} \\\\ \dfrac{x} {xz}+\dfrac{z} {xz}&=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1} {z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{yz}{y+z}} &=& \mathbf{3} \\\\ \dfrac{y+z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)+(2)-(3): & \dfrac{1}{y}+\dfrac{1}{x} + \dfrac{1} {z}+\dfrac{1}{x}- \left( \dfrac{1}{z}+\dfrac{1}{y} \right) &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{x} &=& 6 + 3- 2 \\\\ & \dfrac{12}{x} &=& 7 \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{y=\ ?} \\ \hline (1)-(2)+(3): & \dfrac{1}{y}+\dfrac{1}{x} - \left(\dfrac{1} {z}+\dfrac{1}{x}\right)+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{y} &=& 6 - 3+ 2 \\\\ & \dfrac{12}{y} &=& 5 \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{z=\ ?} \\ \hline -(1)+(2)+(3): & - \left( \dfrac{1}{y}+\dfrac{1}{x} \right) + \dfrac{1} {z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{z} &=& -6 + 3+ 2 \\\\ & \dfrac{12}{z} &=& -1 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array}\)
