Problem 9 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 10, solve for \(x\). $$ \left|\frac{x+2}{x-2}\right|=5 $$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \) and \( x = \frac{4}{3} \).

Step 1 - Understand the absolute value equation

For the equation \ \ \ \ \ |f(x)| = c, the expression inside the absolute value can be either \ c \ or \ -c. Therefore, rewrite the given equation without the absolute value as two separate cases: \ \( \frac{x+2}{x-2} = 5 \) and \ \( \frac{x+2}{x-2} = -5 \).

Step 2 - Solve the first equation

Consider the equation \( \frac{x+2}{x-2} = 5 \). Cross-multiply to solve for \(x\): \ \( x + 2 = 5(x - 2) \). \ This simplifies to \( x + 2 = 5x - 10 \). \ Bring all terms involving \( x \) to one side: \ \( 2 + 10 = 5x - x \) which gives \( 12 = 4x \). \ Therefore, \( x = 3 \).

Step 3 - Solve the second equation

Consider the equation \( \frac{x+2}{x-2} = -5 \). Cross-multiply to solve for \( x \) again: \ \( x + 2 = -5(x - 2) \). \ This simplifies to \( x + 2 = -5x + 10 \). \ Bring all terms involving \( x \) to one side: \ \( x + 5x + 2 = 10 \) which gives \( 6x + 2 = 10 \). \ Therefore, \( 6x = 8 \) and \( x = \frac{4}{3} \).

Step 4 - Verify the solutions

Insert both solutions back into the original equation to ensure they are valid. \ For \( x = 3 \): \ \( \left| \frac{3+2}{3-2} \right| = \left|5\right| = 5 \), which is true. \ For \( x = \frac{4}{3} \): \ \( \left| \frac{\frac{4}{3} + 2}{\frac{4}{3} - 2} \right| = \left| \frac{\frac{10}{3}}{-\frac{2}{3}} \right| = \left| -5 \right| = 5 \), which is also true. Both solutions are correct.

Key Concepts

Solving Equations

Solving equations means finding the value of the unknown variable that makes the equation true. In this exercise, the given equation involves absolute values, which require special handling. The objective is to determine the value of \( x \) that satisfies the equation \( \left| \frac{x+2}{x-2} \right| = 5 \).

Absolute value equations like \( |f(x)| = c \) mean that \( f(x) \) can be either \( c \) or \( -c \). This results in two separate equations to solve:

• \( \frac{x+2}{x-2} = 5 \)
• \( \frac{x+2}{x-2} = -5 \)

Each equation is then solved to find potential solutions for \( x \).

To solve these equations, we need to isolate \( x \). Cross-multiplication becomes a useful tool here, helping to clear the fractions.

Cross-Multiplication

Cross-multiplication is a method to solve equations involving fractions. By multiplying the numerator of one fraction by the denominator of the other, we eliminate the fraction and simplify the equation.

For the first equation \( \frac{x+2}{x-2} = 5 \), cross-multiplying gives us:

• \( x + 2 = 5(x - 2) \)

Expanding and rearranging terms results in:

• \( x + 2 = 5x - 10 \)

Bringing like terms together, we get:

\( 12 = 4x \)

Therefore,

• \( x = 3 \).

Similarly, for the second equation \( \frac{x+2}{x-2} = -5 \), cross-multiplying yields:

• \( x + 2 = -5(x - 2) \)

Expanding and rearranging terms results in:

• \( x + 2 = -5x + 10 \)

Bringing like terms together, we get:

\( 6x = 8 \)

Therefore,

• \( x = \frac{4}{3} \).

This step-by-step process helps break down complex equations into manageable steps.

Verification of Solutions

Verification is the final step in solving equations. It ensures that the values we found for \( x \) are correct by substituting them back into the original equation.

For the first solution \( x = 3 \):

• Substitute \( x = 3 \) into \( \left| \frac{3+2}{3-2} \right| \)
• This simplifies to \( \left| 5 \right| = 5 \), which is correct.

For the second solution \( x = \frac{4}{3} \):

• Substitute \( x = \frac{4}{3} \) into \( \left| \frac{\frac{4}{3} + 2}{\frac{4}{3} - 2} \right| \)
• This simplifies to \( \left| \frac{\frac{10}{3}}{-\frac{2}{3}} \right| \), which is \( \left| -5 \right| = 5 \), also correct.

This verification step is crucial because it confirms that the solutions satisfy the original equation. It's a good practice to always check your solutions to avoid errors.