91Ó°ÊÓ

A system of equations consists of two or more equations that share the same set of unknowns. In the given exercise, we have a system of two equations involving variables \(x\) and \(y\): \[\begin{align*}1. & \quad x^2 + y^2 = 13 \2. & \quad x^2 - y^2 = 5 \end{align*}\] The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. This means we're looking for a pair of numbers which, when plugged into both equations, make both sides of each equation equal. Systems of equations can often pose challenges due to multiple variables. However, using methods like substitution, graphing, or elimination, we can solve them efficiently. In this exercise, we employ the elimination method, sometimes called the addition method, to find the solutions.

The elimination method is a powerful tool for solving systems of equations. It works by removing one variable entirely, allowing us to solve for the other one. This method is sometimes referred to as the addition method because it often involves adding or subtracting equations to cancel out a variable. In our exercise, we are tasked with eliminating one of the variables, \(x^2\), by subtracting the second equation from the first: \[\begin{align*}(x^2 + y^2) - (x^2 - y^2) = 13 - 5 \end{align*}\] This step not only removes \(x^2\) but also simplifies the system to a single equation: \[2y^2 = 8\]By clearing away one variable, the elimination method turns a complex problem into a simpler one-variable equation that can be solved with ease.

Solving quadratic equations is a key skill when tackling systems of equations like the one we have. In this case, after elimination, we're left with the quadratic equation \(2y^2 = 8\). To solve for \(y\): 1. **Simplify the equation** by dividing both sides by 2: \[y^2 = 4\]2. **Take the square root** of both sides to find values for \(y\): \[y = \pm 2\] The square root introduces \(\pm\) because both 2 and -2 satisfy the equation \((\pm 2)^2 = 4\). Next, we substitute these values one at a time back into either of the original equations to find corresponding \(x\) values. For example, using \(y = 2\) in the second equation: \[x^2 - 4 = 5 \rightarrow x^2 = 9\] Then, taking the square root of both sides gives \(x = \pm 3\). The same process is repeated for \(y = -2\), leading to the same \(x\) values. Thus, the complete set of solutions for the system includes the pairs \((3, 2), (-3, 2), (3, -2), (-3, -2)\). Quadratic equations often arise in systems with non-linear components, making the ability to solve them crucial.