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Solution

$ \alpha(1 - \alpha) = 2\lambda $ $ \lambda = \frac{\alpha - \alpha^2}{2} \quad ...(5)$ Use (4): $ \alpha\left(\frac{10}{3} - \alpha\right) = 9\lambda $ Substitute (5): $ \alpha\left(\frac{10}{3} - \alpha\right) = 9 \cdot \frac{\alpha - \alpha^2}{2} $ Multiply by 6: $ 6\alpha\left(\frac{10}{3} - \alpha\right) = 27(\alpha - \alpha^2) $ $ 20\alpha - 6\alpha^2 = 27\alpha - 27\alpha^2 $ $ 0 = 7\alpha - 21\alpha^2 $ $ \alpha(1 - 3\alpha) = 0 $ Since $ \lambda \ne 0 \Rightarrow \alpha \ne 0 $ So, $ \alpha = \frac{1}{3} $ From (1): $ \beta = 1 - \frac{1}{3} = \frac{2}{3} $ From (5): $ \lambda = \frac{\frac{1}{3} - \frac{1}{9}}{2} = \frac{\frac{2}{9}}{2} = \frac{1}{9} $ Now, $ \frac{\beta \gamma}{\lambda} $ First find $ \gamma $: $ \gamma = \frac{10}{3} - \frac{1}{3} = 3 $ So, $ \beta \gamma = \frac{2}{3} \cdot 3 = 2 $ Thus, $ \frac{\beta \gamma}{\lambda} = \frac{2}{1/9} = 18 $