\( \left. \begin{array} { l l } { 1.6 } & { 3 k ( 1 - k ) + 5 ( k + 1 ) = 0 } \\ { 1.8 } & { b ( b + 5 ) = 6 } \end{array} \right. \)

**Step 1. Solve the first equation: \(3k(1 - k) + 5(k + 1) = 0\)** 1. Expand the products: \[ 3k(1 - k) = 3k - 3k^2 \quad \text{and} \quad 5(k + 1) = 5k + 5. \] 2. Write the equation: \[ 3k - 3k^2 + 5k + 5 = 0. \] 3. Combine like terms: \[ -3k^2 + 8k + 5 = 0. \] 4. Multiply the equation by \(-1\) to simplify: \[ 3k^2 - 8k - 5 = 0. \] 5. Apply the quadratic formula, where \( a = 3 \), \( b = -8 \), and \( c = -5 \): \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}. \] 6. Compute the discriminant: \[ (-8)^2 - 4(3)(-5) = 64 + 60 = 124. \] 7. Write the solutions: \[ k = \frac{8 \pm \sqrt{124}}{6}. \] 8. Simplify \(\sqrt{124}\): \[ \sqrt{124} = 2\sqrt{31}. \] 9. Thus, the solutions for \( k \) are: \[ k = \frac{8 \pm 2\sqrt{31}}{6} = \frac{4 \pm \sqrt{31}}{3}. \] --- **Step 2. Solve the second equation: \(b(b + 5) = 6\)** 1. Expand the product: \[ b^2 + 5b = 6. \] 2. Rearranging gives a quadratic equation: \[ b^2 + 5b - 6 = 0. \] 3. Factor the quadratic equation: \[ b^2 + 5b - 6 = (b + 6)(b - 1) = 0. \] 4. Set each factor equal to zero: - \(b + 6 = 0 \Rightarrow b = -6\). - \(b - 1 = 0 \Rightarrow b = 1\). --- **Final Answers:** - For \( k \): \[ k = \frac{4+\sqrt{31}}{3} \quad \text{or} \quad k = \frac{4-\sqrt{31}}{3}. \] - For \( b \): \[ b = -6 \quad \text{or} \quad b = 1. \]