Solve each system: \[ \left\{\begin{array}{l}y=x^{2}+6 x \\ y=4 x\end{array}\right. \] \( \left\{\begin{array}{l}y=x^{2}+8 x+16 \\ x-y=-4\end{array}\right. \)

To solve the first system of equations: \[ \begin{aligned} 1. & \quad y = x^2 + 6x \\ 2. & \quad y = 4x \end{aligned} \] Set the two equations equal to each other: \[ 4x = x^2 + 6x \] Rearranging gives: \[ 0 = x^2 + 6x - 4x \implies 0 = x^2 + 2x \implies 0 = x(x + 2) \] So, we have \(x = 0\) and \(x = -2\). Now, we find \(y\) for each \(x\): - For \(x = 0\): \[ y = 4(0) = 0 \quad \Rightarrow \quad (0, 0) \] - For \(x = -2\): \[ y = 4(-2) = -8 \quad \Rightarrow \quad (-2, -8) \] Thus, the solutions for the first system are \((0, 0)\) and \((-2, -8)\). Now, for the second system of equations: \[ \begin{aligned} 1. & \quad y = x^2 + 8x + 16 \\ 2. & \quad x - y = -4 \quad \text{(or } y = x + 4\text{)} \end{aligned} \] Set the two equations equal to each other: \[ x + 4 = x^2 + 8x + 16 \] Rearranging gives: \[ 0 = x^2 + 8x + 16 - x - 4 = x^2 + 7x + 12 \] Factoring, we have: \[ 0 = (x + 3)(x + 4) \] So, \(x = -3\) and \(x = -4\). Now, find \(y\) for each \(x\): - For \(x = -3\): \[ y = -3 + 4 = 1 \quad \Rightarrow \quad (-3, 1) \] - For \(x = -4\): \[ y = -4 + 4 = 0 \quad \Rightarrow \quad (-4, 0) \] Thus, the solutions for the second system are \((-3, 1)\) and \((-4, 0)\). In summary, the solutions are: 1. First system: \((0, 0)\) and \((-2, -8)\) 2. Second system: \((-3, 1)\) and \((-4, 0)\)