Solve for \( x \) and \( y \) simulameonsly \( 4-y+2=0 \) and \( y=-x^{2}+2 x+8 \)

We begin with the system of equations: \[ 4 - y + 2 = 0 \] \[ y = -x^2 + 2x + 8 \] **Step 1: Solve the first equation for \( y \).** Combine like terms in the first equation: \[ 4 + 2 - y = 0 \quad \Rightarrow \quad 6 - y = 0 \] Solve for \( y \): \[ y = 6 \] **Step 2: Substitute \( y = 6 \) into the second equation.** Substitute into: \[ 6 = -x^2 + 2x + 8 \] Rearrange the equation: \[ -x^2 + 2x + 8 - 6 = 0 \quad \Rightarrow \quad -x^2 + 2x + 2 = 0 \] Multiply the entire equation by \(-1\) to simplify: \[ x^2 - 2x - 2 = 0 \] **Step 3: Solve the quadratic equation for \( x \).** The quadratic equation is: \[ x^2 - 2x - 2 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = 1 \), \( b = -2 \), and \( c = -2 \). Substitute these values: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} \] Simplify \(\sqrt{12}\): \[ \sqrt{12} = 2\sqrt{3} \] Thus: \[ x = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \] **Final Answer:** The solutions to the system are: \[ x = 1 + \sqrt{3} \quad \text{or} \quad x = 1 - \sqrt{3}, \quad \text{and} \quad y = 6. \]