Model - Solve Linear-Quadratic Systems of Equations Using Elimination: What are the solutions to the system of equations below? \[ \begin{array}{l} y=x^{2}-6 x+8 \\ y=2 x-4 \end{array} \]

To solve the system of equations using the elimination method, you start by setting the two expressions for \(y\) equal to each other since they both represent \(y\): \[ x^2 - 6x + 8 = 2x - 4 \] Rearranging gives: \[ x^2 - 8x + 12 = 0 \] This factors into: \[ (x - 6)(x - 2) = 0 \] So, the solutions for \(x\) are \(x = 6\) and \(x = 2\). To find corresponding \(y\) values, substitute these \(x\) values back into either equation, typically the linear one for simplicity. For \(x = 6\): \[ y = 2(6) - 4 = 8 \] For \(x = 2\): \[ y = 2(2) - 4 = 0 \] Thus, the solutions are \((6, 8)\) and \((2, 0)\). The method used frequently shows that elimination can be both effective and fun, especially when items start factoring nicely, like getting a perfect quadratic! Plus, there’s something satisfying about locating points (where they intersect on the graph) as they tell a story of where our equations come together. If you’re tackling quadratic equations, keep in mind that factoring isn’t your only friend. The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), can swoop in to save the day when factoring seems daunting! And remember, a simple mistake in arithmetic can throw off your results entirely, so double-checking your work is always a smart move!