Problem 11 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of nonlinear equations using elimination. $$ \begin{array}{l} 4 x^{2}-9 y^{2}=36 \\ 4 x^{2}+9 y^{2}=36 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We have a system of two nonlinear equations:\[ 4x^2 - 9y^2 = 36 \]\[ 4x^2 + 9y^2 = 36 \]The goal is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Adding the Equations

Add the two equations to eliminate \(y^2\):\[(4x^2 - 9y^2) + (4x^2 + 9y^2) = 36 + 36\]This simplifies to:\[8x^2 = 72\]

Step 3: Solving for x

Solve \(8x^2 = 72\) for \(x^2\):\[8x^2 = 72 \x^2 = \frac{72}{8} = 9\]Take the square root of both sides:\[x = \pm 3\]

Step 4: Subtracting the Equations

Subtract the second equation from the first to eliminate \(x^2\):\[(4x^2 - 9y^2) - (4x^2 + 9y^2) = 36 - 36\]This simplifies to:\[-18y^2 = 0\]

Step 5: Solving for y

Solve \(-18y^2 = 0\) for \(y^2\):\[y^2 = 0\]Take the square root:\[y = 0\]

Step 6: Verifying the Solutions

Substitute \(x = 3\) and \(x = -3\) into the original equations to verify:For \(x = 3\):1. \(4(3)^2 - 9(0)^2 = 36\)2. \(4(3)^2 + 9(0)^2 = 36\)Both equations are satisfied.For \(x = -3\):1. \(4(-3)^2 - 9(0)^2 = 36\)2. \(4(-3)^2 + 9(0)^2 = 36\)Both equations are satisfied.

Key Concepts

Elimination Method

The elimination method is a popular technique used to solve systems of equations. This method involves combining equations to eliminate one of the variables. Once a variable is eliminated, the remaining equation, which now involves only one variable, can be solved more easily.
In the given problem:

• We start with two nonlinear equations: \(4x^2 - 9y^2 = 36\) and \(4x^2 + 9y^2 = 36\).
• Adding these equations eliminates \(y^2\), leading to a simpler equation: \(8x^2 = 72\).
• Subtracting the second equation from the first eliminates \(x^2\), resulting in \(-18y^2 = 0\).
This strategic approach efficiently provides isolated equations for each variable, allowing us to solve them independently.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving a system means finding a set of variable values that satisfies all equations in the system simultaneously.
There are various methods for solving systems, such as:

• Graphing
• Substitution
• Elimination

For nonlinear systems like the one we have, the elimination method can be particularly effective because it simplifies the problem by reducing the number of variables in each equation. By adding and subtracting equations:

• We obtained simpler quadratic equations focused on one variable at a time.
• This led to finding simple solutions for \(x\) and \(y\) which satisfy both original equations.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, generally expressed as \(ax^2 + bx + c = 0\). In the system we are solving, both equations are quadratic in terms of \(x\) and \(y\).
Here's how we approached these quadratic equations:

• After applying the elimination method, we simplified the quadratic into \(8x^2 = 72\), which can be solved by first dividing through by 8.
• The remaining equation \(x^2 = 9\) gives two potential solutions for \(x\): \(x = 3\) or \(x = -3\).
• The second quadratic \(-18y^2 = 0\) simplifies to \(y^2 = 0\), indicating a unique solution of \(y = 0\).

This demonstrates how powerful quadratic equations are in providing possible values for variables, and when combined with methods like elimination, allows us to solve complex nonlinear systems effectively.