Problem 49 - College Algebra

Question

For the following exercises, find the solutions to the nonlinear equations with two variables. $$\frac{6}{x^{2}}-\frac{1}{y^{2}}=8$$ $$\frac{1}{x^{2}}-\frac{6}{y^{2}}=\frac{1}{8}$$

Step-by-Step Solution

Answer

The solutions are approximately \( x \approx 0.858 \) and \( y \approx 2.192 \).

Step 1: Equation Rearrangement

First, rearrange both equations to make them more manageable. The original equations are:1. \( \frac{6}{x^2} - \frac{1}{y^2} = 8 \)2. \( \frac{1}{x^2} - \frac{6}{y^2} = \frac{1}{8} \)Let's express these equations with a common variable: Let \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \). Thus, we rewrite the equations as follows:1. \( 6u - v = 8 \)2. \( u - 6v = \frac{1}{8} \)

Step 2: Solve the System of Equations

We now have a system of linear equations:1. \( 6u - v = 8 \)2. \( u - 6v = \frac{1}{8} \)Let's solve this system using substitution or elimination. We'll use substitution by expressing \( v \) from the first equation:\( v = 6u - 8 \).Substitute \( v \) in the second equation:\( u - 6(6u - 8) = \frac{1}{8} \)Simplify:\( u - 36u + 48 = \frac{1}{8} \)\( -35u = \frac{1}{8} - 48 \)\( -35u = \frac{1}{8} - \frac{384}{8} \)\( -35u = \frac{-383}{8} \)\( u = \frac{383}{280} \).

Step 3: Find Value of v

Now, substitute \( u = \frac{383}{280} \) back into the expression for \( v \):\( v = 6\left(\frac{383}{280}\right) - 8 \)\( v = \frac{2298}{280} - 8 \)\( v = \frac{2298 - 2240}{280} \)\( v = \frac{58}{280} \)\( v = \frac{29}{140} \).

Step 4: Solve for x and y from u and v

Now we have the values for \( u \) and \( v \):\( u = \frac{383}{280} = \frac{1}{x^2} \)\( v = \frac{29}{140} = \frac{1}{y^2} \)Solve for \( x \):\( x^2 = \frac{280}{383} \)\( x = \sqrt{\frac{280}{383}} \)\( x \approx 0.858 \, \text{(approximate)} \)Solve for \( y \):\( y^2 = \frac{140}{29} \)\( y = \sqrt{\frac{140}{29}} \)\( y \approx 2.192 \, \text{(approximate)} \).

Step 5: Verify the Solution

Verify whether the solutions satisfy the original equations:First Equation:\[ \frac{6}{0.858^2} - \frac{1}{2.192^2} \approx 7.995 \] (close to 8)Second Equation:\[ \frac{1}{0.858^2} - \frac{6}{2.192^2} \approx 0.125 \] (close to \(\frac{1}{8}\))Both approximate solutions confirm the validity of the calculated values for \( x \) and \( y \).

Key Concepts

System of Linear Equations

A system of linear equations involves multiple equations that are linear in nature. Each equation in the system contains one or more variables, and all equations are solved simultaneously to find common solutions for these variables.
Linear equations have the standard form of \( ax + by = c \).
In our original exercise, the nonlinear equations were transformed into linear equations by introducing new variables, \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \).
This transformation allowed the nonlinear equation to be treated as a system of linear equations, specifically:

• \( 6u - v = 8 \)
• \( u - 6v = \frac{1}{8} \)

This makes it easier to apply techniques like substitution and elimination to find solutions.

Equation Rearrangement

Rearranging equations is a crucial first step when dealing with complex equations. This involves transforming the original equations into a different form that is simpler to solve.
For the given nonlinear equations:

• \( \frac{6}{x^2} - \frac{1}{y^2} = 8 \)
• \( \frac{1}{x^2} - \frac{6}{y^2} = \frac{1}{8} \)

We use substitution to introduce variables \( u = \frac{1}{x^2} \) and \( v = \frac{1}{y^2} \), which simplifies the process.
Changing the variables in this way allows us to rewrite the equations as linear forms:

• \( 6u - v = 8 \)
• \( u - 6v = \frac{1}{8} \)

This transformation is the key to making further calculations more straightforward.

Substitution Method

The substitution method is a technique used to solve a system of equations, especially when the equations have already been rearranged into a more suitable form.
For our transformed system of equations:

• \( 6u - v = 8 \)
• \( u - 6v = \frac{1}{8} \)

We solved it by expressing one variable, \( v \), in terms of the other, \( u \), using the first equation: \( v = 6u - 8 \).
This expression was then substituted into the second equation, which allowed us to solve for \( u \) first.
Once \( u = \frac{383}{280} \) was found, the value was substituted back to find \( v = \frac{29}{140} \).
The substitution method is powerful as it reduces the system to a single-variable equation.

Verification of Solutions

After computing potential solutions for a system of equations, it is essential to verify them by plugging the values back into the original equations.
Verification ensures that the solutions do not only satisfy the transformed equations but also the initial nonlinear problem in the context of the exercise:

• The values \( x = \sqrt{\frac{280}{383}} \) and \( y = \sqrt{\frac{140}{29}} \) were calculated from \( u \) and \( v \).
• Substituting back into the original equations demonstrated accuracy:
  • The first equation was approximately satisfied: \( \frac{6}{0.858^2} - \frac{1}{2.192^2} \approx 8 \).
  • The second equation was also satisfied within a reasonable margin: \( \frac{1}{0.858^2} - \frac{6}{2.192^2} \approx \frac{1}{8} \).

Verification not just supports the validity of the solution but boosts confidence that the calculations were performed correctly and comprehensively.