Problem 68 - A Graphical Approach to Precalculus with Limits

Question

Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} &2 x^{2}+3 y^{2}=5\\\ &3 x^{2}-4 y^{2}=-1 \end{aligned}$$

Step-by-Step Solution

Answer

The real solutions are \((1, 1)\), \((1, -1)\), \((-1, 1)\), and \((-1, -1)\).

Step 1: Multiply Equations to Eliminate y

To eliminate one of the variables, \( y^2 \), we will add multiples of the equations together. Multiply the first equation by 4 and the second by 3 to align the coefficients of \( y^2 \):\[4(2x^2 + 3y^2) = 4 \times 5 \3(3x^2 - 4y^2) = 3 \times (-1)\]which results in:\[8x^2 + 12y^2 = 20 \9x^2 - 12y^2 = -3\]

Step 2: Add Equations to Eliminate y

Now add the modified equations to eliminate \( y^2 \):\[(8x^2 + 12y^2) + (9x^2 - 12y^2) = 20 + (-3)\]This simplifies to:\[17x^2 = 17\]Divide both sides by 17:\[x^2 = 1\]Therefore, \( x = 1 \) or \( x = -1 \).

Step 3: Substitute x into Original Equations

Now solve for \( y \). Substitute \( x = 1 \) and \( x = -1 \) back into one of the original equations, say \( 2x^2 + 3y^2 = 5 \). Substituting \( x = 1 \):\[2(1)^2 + 3y^2 = 5 \2 + 3y^2 = 5 \3y^2 = 3 \y^2 = 1\]Thus, \( y = 1 \) or \( y = -1 \). Repeat for \( x = -1 \).

Step 4: Derive Solutions

With \( x = 1 \), we found \( y = 1 \) and \( y = -1 \). Check for \( x = -1 \):- Substitute \( x = -1 \) in \( 2x^2 + 3y^2 = 5 \):\[2(-1)^2 + 3y^2 = 5 \2 + 3y^2 = 5 \3y^2 = 3 \y^2 = 1\]Thus, \( y = 1 \) or \( y = -1 \).Consequently, the solutions are \( (1, 1) \), \( (1, -1) \), \( (-1, 1) \), \( (-1, -1) \).

Key Concepts

Analytical Solutions

Analytical solutions involve finding exact values for the unknowns in equations. In our exercise, we're dealing with a system of nonlinear equations, which means you have polynomials of degree greater than one. The term "analytical" signifies that we are getting exact solutions using algebraic methods as opposed to numerical approximations. Let's break down these steps:

• First, identify the type of system. Are the equations linear or nonlinear? In this example, they are nonlinear due to the presence of squared terms like \( x^2 \) and \( y^2 \).
• Determine a method for elimination or substitution. We'll eliminate one of the variables, \( y^2 \) in this case, by manipulating the coefficients with multiplication.
• Combine equations to isolate and solve for one variable. Once you've isolated it, you'll substitute this back into the original equations to find the other variable.

By applying these techniques, you analytically arrive at exact values for \( x \) and \( y \).

Real Solutions

Real solutions are specific types of answers that do not have imaginary numbers. In the step-by-step solution provided, all solutions were real because they didn't involve the square root of a negative number. This means every solution we find will have real-number coordinates.Let's explore why real solutions are important:

• Real solutions are often needed in practical, real-world contexts, where imaginary numbers don’t make sense.
• To ensure solutions are real, check that each step retains variables under even roots only when non-negative.
• For our system, the process of elimination by adding equations led to a clean, real solution \( x^2 = 1 \), which easily translates to real values for \( x \) and \( y \).

This is essential for problems looking to model realistic scenarios.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to solve for unknowns. This tool is critical in solving systems of nonlinear equations. Consider the following stratagems:

• Multiplying equations: This realignment of variables often makes solving easier by equating coefficients.
• Adding or subtracting equations: This removes one variable entirely from the equations, helping isolate another.
• Substitution: Once a variable is isolated, substituting it back is essential to solve for the remaining unknown.

These steps required careful algebraic moves, like distributing multiplication across terms correctly, and keeping terms balanced by performing the same operation on both sides of the equations. Practicing algebraic manipulation refines your skills to approach a variety of algebraic systems efficiently. This process is what facilitated the conversion of our original complex system to simpler components, ultimately leading to the solution.