Try it: Use elimination to solve each system of equations: \[ \begin{array}{l|l} y=-x+4 & y=-x^{2}+4 x+2 \\ y=x^{2}-2 & y=2-x \end{array} \]

To solve the first system using elimination, we can set the equations equal to each other since both equal \(y\): 1. From the first equation: \(y = -x + 4\) 2. From the second equation: \(y = -x^2 + 4x + 2\) Setting them equal gives: \[-x + 4 = -x^2 + 4x + 2\] Rearranging this results in: \[x^2 - 5x + 2 = 0\] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=1, b=-5, c=2\): \[x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}\] \[x = \frac{5 \pm \sqrt{25 - 8}}{2}\] \[x = \frac{5 \pm \sqrt{17}}{2}\] Now, we can substitute these \(x\) values back into \(y = -x + 4\) to find \(y\). For the second system, equate the two expressions for \(y\): 1. From \(y = x^2 - 2\) 2. From \(y = 2 - x\) Set them equal: \[x^2 - 2 = 2 - x\] Rearranging gives us: \[x^2 + x - 4 = 0\] Again, applying the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Where \(a=1, b=1, c=-4\): \[x = \frac{-1 \pm \sqrt{(1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\] \[x = \frac{-1 \pm \sqrt{1 + 16}}{2}\] \[x = \frac{-1 \pm \sqrt{17}}{2}\] Now substitute these \(x\) values back into \(y = 2 - x\) to find \(y\).