91Ó°ÊÓ

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They frequently appear when solving problems involving parabolic shapes and can often occur in solving systems of nonlinear equations, as seen in the given exercise. To solve a quadratic equation, you generally want to isolate the variable squared term. In this exercise, once we simplified to \(2x^2 = 18\), we divided by 2 to get \(x^2 = 9\). Next, by taking the square root of both sides, we find the possible values of x, \(x = \pm 3\). It's essential to check both the positive and negative roots to ensure all solutions are found. This understanding can be critical because missing a potential solution could render the entire solution incomplete.

Eliminating variables is a strategy used to simplify systems of equations to make them easier to solve. In the given exercise, we started with two equations: \(x^2 + y^2 = 9\) and \(x^2 - y^2 = 9\). The key step here was adding the two equations together. By doing so, the \(y^2\) terms canceled each other out: \( (x^2 + y^2) + (x^2 - y^2) = 18\), simplifying to \(2x^2 = 18\). This technique reduced the problem from solving a system of two variables to solving for a single variable, greatly simplifying the task.

The substitution method involves solving one equation for one variable and then substituting that expression into another equation. In our problem, after finding \(x = \pm 3\), we substituted these values back into one of the original equations to find y. For example, using \(x = 3\), we substituted into \(3^2 + y^2 = 9\), resulting in \(9 + y^2 = 9\). Solving this gave us \( y^2 = 0 \) and subsequently \(y = 0 \). By repeating this process for \(-3\), we confirmed that both values of x gave us the same result for y. This method ensures both variables are accounted for accurately and serves as an essential tool for solving systems of nonlinear equations.