Problem 25 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} 3 x^{2}+y^{2}=9 \\ 3 x^{2}-y^{2}=9 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Step 1: Understanding the System

We have a system of nonlinear equations: \(3x^2 + y^2 = 9\) and \(3x^2 - y^2 = 9\). Both equations look similar, so we can use elimination to solve for one variable.

Step 2: Eliminate \(y^2\)

To eliminate \(y^2\), subtract the second equation from the first. This gives:\[(3x^2 + y^2) - (3x^2 - y^2) = 9 - 9\]which simplifies to \(2y^2 = 0\).

Step 3: Solve for \(y\)

From \(2y^2 = 0\), we divide both sides by 2 to get \(y^2 = 0\). Therefore, \(y = 0\).

Step 4: Substitute \(y = 0\) into one equation

Substitute \(y = 0\) into the first equation: \[3x^2 + 0^2 = 9\] which simplifies to \(3x^2 = 9\).

Step 5: Solve for \(x\)

Divide both sides by 3:\[x^2 = 3\].Taking the square root of both sides, we find:\[x = \pm\sqrt{3}\].

Step 6: Write the Solution Pair

With \(y = 0\), the solutions for \(x\) are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\). Thus, the solution pairs are \((x, y) = (\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Key Concepts

Understanding Systems of Equations

A system of equations is a set of two or more equations that share the same set of variables. When you solve a system of equations, you are looking for the values of these variables that satisfy each equation in the system. In the context of this exercise, we have two polynomial equations involving the variables \(x\) and \(y\). Solving these equations simultaneously will give us the solutions where both equations are true at the same time. Systems of equations can be classified into:

• Linear systems, where each equation is linear.
• Nonlinear systems, where at least one equation is nonlinear, such as having variables raised to a power > 1.

In our exercise, the equations are nonlinear due to the quadratic terms \(x^2\) and \(y^2\). When faced with such systems, it is useful to utilize strategic methods like elimination or substitution to find solutions.

Exploring the Elimination Method

The elimination method aims to simplify a system of equations by removing one variable, which makes it easier to solve for the remaining variables. This is achieved by adding or subtracting equations to cancel out one of the variables. In our exercise, the similarity of the two equations hints that elimination could be effective.Let's break down the process:

• We subtract the second equation from the first: \((3x^2 + y^2) - (3x^2 - y^2) = 9 - 9\).
• This simplifies to \(2y^2 = 0\).
• Solving \(2y^2 = 0\) directly gives us \(y^2 = 0\), leading to \(y = 0\).

The elimination method swiftly determined the value of \(y\) by effectively removing it, paving the way to focus on solving for \(x\) in a simpler equation.

Harnessing the Power of Square Roots

Taking square roots is a key technique for solving equations involving squared terms. When you solve for \(x\) or \(y\) in quadratic equations, you will often end up needing to take a square root.Once we have \(y = 0\), we substitute back into an equation:

• Substitute to get \(3x^2 = 9\).
• Divide each side by 3, leading to \(x^2 = 3\).
• Take the square root: \(x = \pm\sqrt{3}\).

This technique is useful because squares and square roots are inverse operations. Keep in mind that taking square roots can yield both positive and negative solutions due to the dual nature of roots, as seen in \(x = \pm\sqrt{3}\).

The Result: Solution Pairs

Solution pairs represent the combinations of \(x\) and \(y\) that satisfy the entire system of equations simultaneously. After solving the equations, the solution pairs provide you with the comprehensive list of values that work for both equations.For our specific system:

• We found \(y = 0\).
• Solving for \(x\), we have \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

Thus, the solution pairs are:

• \((x, y) = (\sqrt{3}, 0)\)
• \((x, y) = (-\sqrt{3}, 0)\)

These pairs represent all possible scenarios where the given nonlinear system holds true, effectively solving the original problem.