λ=μ

question:

\(\lambda=\mu \\ \dfrac{1}{2}\lambda=1-\mu \\ 1-\lambda=\dfrac{1}{2}\mu \\ \lambda=\mu=?\)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{1}{2}\lambda &=& 1-\mu \\ & \dfrac{1}{2}\lambda+\mu &=& 1 \quad | \quad \times 4 \\ & \mathbf{2\lambda+4\mu} &=& \mathbf{4} \\\\ (2) & 1-\lambda &=& \dfrac{1}{2}\mu \\ & \lambda +\dfrac{1}{2}\mu &=& 1 \quad | \quad \times 2 \\ & \mathbf{2\lambda + \mu} &=& \mathbf{2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{2\lambda+4\mu} &=& \mathbf{4} \\ (2) & \mathbf{2\lambda + \mu} &=& \mathbf{2} \\ \hline (1)-(2): & 4\mu- \mu &=& 4-2 \\ & 3\mu &=& 2 \\ & \mathbf{ \mu } &=& \mathbf{\dfrac{2}{3}} \\ \hline \end{array} \)

\(\lambda=\mu=\mathbf{\dfrac{2}{3}}\)

Could also simply do the following:

\(\text{If }\lambda=\mu\text{ then }\frac{1}{2}\lambda=1-\mu\text{ becomes }\frac{1}{2}\lambda=1-\lambda\\ \text{ so } \frac{3}{2}\lambda=1\text{ or }\lambda=\frac{2}{3}\)