Problem 93 - College Algebra with Modeling and Visualization

Question

Use elimination to solve the nonlinear system of equations. $$ \begin{aligned} &x^{2}+y=12\\\ &x^{2}-y=6 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((3, 3)\) and \((-3, 3)\).

Step 1: Set up the Elimination

We will use the elimination method to solve the system of equations. Rewrite the given equations:1. \(x^2 + y = 12\)2. \(x^2 - y = 6\)We aim to eliminate \(y\) by adding these two equations.

Step 2: Add the Equations

Add the two equations together to eliminate \(y\):\[(x^2 + y) + (x^2 - y) = 12 + 6\]This simplifies to:\[2x^2 = 18\]

Step 3: Solve for \(x\)

To find \(x\), divide both sides of the equation by 2:\[x^2 = 9\]Then, take the square root of both sides:\[x = 3 \text{ or } x = -3\]

Step 4: Substitute \(x\) back into one of the original equations

Substitute the values of \(x\) back into one of the original equations to find \(y\). Using equation 1, \(x^2 + y = 12\):- If \(x = 3\): \[ 3^2 + y = 12 \ 9 + y = 12 \ y = 3 \]- If \(x = -3\): \[ (-3)^2 + y = 12 \ 9 + y = 12 \ y = 3 \]

Step 5: Write the Solution

The solutions to the system are the pairs \((x, y)\). From the calculations, we have two solutions: \((3, 3)\) and \((-3, 3)\).

Key Concepts

Elimination Method

The elimination method is a popular technique used to solve systems of equations. It involves adding or subtracting equations in such a way that one of the variables is eliminated, making it easier to solve for the remaining variable(s). This method is particularly handy for systems with linear equations, but it can also be applied to nonlinear equations as seen in the given problem.

For nonlinear systems, like the one in the exercise, the elimination method can still be effective. In our example, we have two equations with the variable \(y\) set up such that by adding the equations, \(y\) disappears, leaving us with only \(x^2\) to solve. The key here is to carefully set up the equations and strategically choose to add or subtract to eliminate one variable. This streamlines the process of finding solutions for the remaining variables.

Solving Equations

Solving equations involves finding the values of the variables that satisfy the equation(s). In a system of equations, the solution is typically a set of values that work for all equations simultaneously. When solving, it's crucial to track each variable and ensure that the operations performed are valid for all involved equations.

For our nonlinear system, after eliminating \(y\), we are left with the simplified equation \(2x^2 = 18\). Solving this involves basic algebraic manipulations: dividing through by 2 to isolate \(x^2\), and then taking the square root to solve for \(x\). This yields \(x = 3\) or \(x = -3\). Both solutions must then be substituted back into one of the original equations to find the corresponding \(y\) values. Solving the system effectively means verifying each potential solution so that it checks out in all original equations.

Algebraic Solutions

Algebraic solutions refer to finding the values of variables by using the rules and operations of algebra. These can be simple equations or, as in our case, involve a system of equations that may include polynomial expressions.

In this example, it is required to apply algebraic methods to find the intersection points of the two given equations. Once the variable \(x\) is isolated and values are determined as \(3\) and \(-3\), we proceed to find \(y\) by back-substituting into \(x^2 + y = 12\). Both values of \(x\) result in \(y\) being \(3\). Thus, the algebraic solutions to the original system are the pairs \((3, 3)\) and \((-3, 3)\).

These solutions exemplify how algebraic reasoning and systematic solving techniques can bring clarity to what appears at first to be a complex nonlinear system.