Problem 49 - College Algebra

Question

Find the solutions to the nonlinear equations with two variables. $$ \begin{array}{l} \frac{6}{x^{2}}-\frac{1}{y^{2}}=8 \\ \frac{1}{x^{2}}-\frac{6}{y^{2}}=\frac{1}{8} \end{array} $$

Step-by-Step Solution

Answer

Solutions are approximately \( x \approx 0.85 \) and \( y \approx 2.19 \).

Step 1: Identify Variables as Substitutions

Introduce new variables to simplify the equations. Let \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \). This means the system of equations becomes: \( 6u - v = 8 \) and \( u - 6v = \frac{1}{8} \).

Step 2: Solve the Linear System

We now deal with a system of linear equations in terms of \( u \) and \( v \): \( 6u - v = 8 \) and \( u - 6v = \frac{1}{8} \). Use the substitution or elimination method to solve this. Solve the first for \( v \), giving \( v = 6u - 8 \). Substitute into the second equation to get \( u - 6(6u - 8) = \frac{1}{8} \).

Step 3: Simplify and Solve for u

Substitute \( v = 6u - 8 \) in the second equation: \( u - 36u + 48 = \frac{1}{8} \). Simplify it to: \(-35u + 48 = \frac{1}{8}\). Isolate \( u \) by subtracting 48 from both sides, resulting in \(-35u = \frac{1}{8} - 48\).

Step 4: Calculate the Exact Value of u

Calculate \( 48 = \frac{384}{8} \) and solve \( -35u = \frac{1}{8} - \frac{384}{8} = -\frac{383}{8} \). Therefore, \( u = \frac{383}{8 \times 35} \).

Step 5: Solve for v Using u's Value

Substitute \( u \) back into \( v = 6u - 8 \). Calculate \( v = 6\left( \frac{383}{280} \right) - 8 = \frac{2298}{280} - \frac{2240}{280} = \frac{58}{280} \).

Step 6: Return to Original Variables

We now go back to \( x \) and \( y \). Since \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \). We have \( x^2 = \frac{280}{383} \) and \( y^2 = \frac{280}{58} \). Solve for \( x \) and \( y \) to get \( x = \sqrt{\frac{280}{383}} \) and \( y = \sqrt{\frac{280}{58}} \).

Step 7: Calculate the Approximate Solutions for x and y

Simplify to get \( x \approx 0.85 \) and \( y \approx 2.19 \) (using a calculator to find square roots).

Key Concepts

Substitution Method

The substitution method is a strategic way to solve systems of equations. It involves replacing one variable with an expression derived from another equation. This method is particularly useful when dealing with nonlinear equations, like in the given exercise. By using substitution, complex equations can be transformed into a simpler form, often linear, making them easier to solve.

In the exercise, we simplify the originally nonlinear equations by letting \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \). This substitution converts the system into two linear equations. This new system is straightforward and much easier to manipulate compared to the original nonlinear equations. Substitution can therefore be an elegant tool when encountering complex relationships between variables in equations.

Systems of Equations

A system of equations consists of two or more equations sharing common variables. The challenge is to find values for these variables that satisfy all equations simultaneously. In this exercise, the system involves two equations with the variables \( x \) and \( y \).

These equations initially appear nonlinear due to their reciprocal square terms. However, with the help of substitutions, these equations transform into linear forms involving new variables \( u \) and \( v \).

• Original system: Solved by converting to the standard form.
• Simplified system: Easier to solve due to linearity.

Understanding how to handle systems of equations is crucial in mathematics. It enables you to solve more than just linear scenarios and apply this knowledge to various fields, such as physics, engineering, and computer science.

Linear Equations

Linear equations are equations of the first order. They represent straight lines in graphical interpretation and are among the simplest forms of equations in algebra. In this case, after substitution, the equations \( 6u - v = 8 \) and \( u - 6v = \frac{1}{8} \) are linear.

Solving these linear equations involves straightforward techniques such as graphing, substitution, and elimination. Here, since we've already substituted to find expressions for \( u \) and \( v \), solving these linear equations becomes more a matter of simplification.

• Replace variables: Use some algebra to rearrange expressions.
• Isolate variables: Find values for each in terms of the others.

Linear equations offer a systematic approach to solving problems where relationships are proportional and constant, as demonstrated in the exercise.

Variable Elimination

Variable elimination is a technique used to remove a variable from a system of equations, simplifying the process of finding a solution. By eliminating a variable, you reduce the number of equations needed to solve, which is helpful in systems containing several equations or complex terms.

In the exercise, we eliminated \( v \) by substituting it with \( 6u - 8 \) into the second equation. This step resulted in an equation with a single variable \( u \), which was then solved easily.

• Efficiency: Makes computations more tractable by focusing on fewer variables.
• Simplicity: Simplifies complex or cumbersome expressions.

Through variable elimination, the system is broken down into easily manageable steps, allowing for efficient resolution of algebraic problems.