Problem 67 - A Graphical Approach to College Algebra

Question

Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}+2 y^{2} &=9 \\ 3 x^{2}-4 y^{2} &=27 \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Understanding the System of Equations

We have two equations: \( x^2 + 2y^2 = 9 \) and \( 3x^2 - 4y^2 = 27 \). Our goal is to find the real values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Express One Variable in Terms of the Other

From the first equation, solve for \(x^2\): \( x^2 = 9 - 2y^2 \). Substitute this expression for \(x^2\) in the second equation to have one equation in terms of \(y\).

Step 3: Substitute and Simplify the Equation

Substituting \(x^2 = 9 - 2y^2\) into the second equation gives \(3(9 - 2y^2) - 4y^2 = 27\). Simplify this expression to \(27 - 6y^2 - 4y^2 = 27\), which reduces to \(-10y^2 = 0\).

Step 4: Solve for \(y\)

Solve \(-10y^2 = 0\). Thus, \(y^2 = 0 \) giving \( y = 0 \).

Step 5: Find Corresponding \(x\) Value

With \(y = 0\), substitute back into \(x^2 = 9 - 2y^2 = 9\) to get \(x^2 = 9\). Thus, \(x = \pm 3\).

Step 6: List All Real Solutions

The pairs \((x, y)\) that solve the system are \((3, 0)\) and \((-3, 0)\).

Key Concepts

Analytical Solutions

When solving a nonlinear system of equations, we aim to find solutions that satisfy all equations involved. Such problems can be more complex as they do not form straight lines, rather, they involve curves, circles, ellipses, or parabolas.
Analytical solutions require us to use precise mathematical methods to solve these equations. The advantage of analytical solutions is that they provide exact answers, unlike numerical methods which estimate the solution. To solve these systems analytically, we often:

• Express one variable in terms of another to reduce the number of equations.
• Substitute and manipulate the equations algebraically to find a solution.

This approach was exactly what was done in the original solution when the variable \(x^2\) was expressed in terms of \(y^2\), simplifying the problem to a single equation in one variable. Once this was achieved, solving for \(y\) became straightforward, providing exact real solutions.

Real Solutions

Real solutions refer to values of variables that satisfy the given equations and are real numbers. In the context of this problem, we want to find values of \(x\) and \(y\) which are not complex numbers, and that satisfy both equations simultaneously.
In the step-by-step solution, simplifying the equations led us to solve \(-10y^2 = 0\), indicating \(y^2 = 0\). This straightforwardly gives \(y = 0\) as a real solution, considering that square roots of zero yield only real numbers. Once \(y\) is found, substituting it into the derived expression for \(x^2\) yields two potential values, \(x = 3\) and \(x = -3\).
Thus, the system has real solutions \((3, 0)\) and \((-3, 0)\), both of which are checked to satisfy the original equations, confirming their validity.

Algebraic Manipulation

Algebraic manipulation is a powerful technique in solving systems of equations. It involves rearranging and simplifying equations to make them easier to solve.
In this problem, algebraic manipulation was crucial. First, \(x^2\) was isolated in the first equation and then substituted into the second equation. This step effectively reduced a system of two equations into a single equation in terms of \(y\).
Further simplification was done by expanding and then combining like terms: \(3(9 - 2y^2) - 4y^2\) became \(-10y^2\), a critical step that led to finding \(y = 0\). Substituting \(y = 0\) back into the expression for \(x^2\) provided the possible values for \(x\) that also checked out with the system.
Mastering algebraic manipulation requires practice, as it provides the flexibility to transform equations into simpler forms, making complex problems more tractable.