Problem 31 - The Calculus with Analytic Geometry

Question

In Exercises 29 through 32, solve for \(x\) and use absolute value bars to write the answer. $$ \frac{x-2}{x-4}>\frac{x+2}{x} $$

Step-by-Step Solution

Answer

The solution is \( x eq 0 \) and \( x eq 4 \).

Step 1: Set up the inequality

Start with the given inequality: \( \frac{x-2}{x-4} > \frac{x+2}{x} \).

Step 2: Eliminate the fractions

To eliminate the fractions, find a common denominator, which is \((x)(x-4)\). Multiply both sides by \((x)(x-4)\): \( (x)(x-4) \cdot \frac{x-2}{x-4} > (x)(x-4) \cdot \frac{x+2}{x} \).

Step 3: Simplify the inequality

Simplify both sides of the inequality: Left side: \( x(x-2) \). Right side: \( (x-4)(x+2) \). Thus, the inequality becomes \( x(x-2) > (x-4)(x+2) \).

Step 4: Expand and simplify

Expand both sides: Left side: \( x^2 - 2x \). Right side: \( x^2 + 2x - 4x - 8 \). Thus, \( x^2 - 2x > x^2 - 2x - 8 \).

Step 5: Cancel like terms

Subtract \(x^2\) and \(-2x\) from both sides to isolate \(x\): \(0 > -8\), which is always true. This implies the inequality holds for all values except where the original fractions are undefined.

Step 6: Identify restrictions

The original fractions are undefined at \( x = 0 \) and \( x = 4 \). Exclude these points: \( x eq 0 \) and \( x eq 4 \).

Step 7: Write the final solution

Combine the solution set with the restrictions in terms of absolute value: The solution is \( x eq 0 \) and \( x eq 4 \), or \( x oti\text{-terms}\{0, 4\}. \).

Key Concepts

absolute value equations

Absolute value equations involve the use of absolute value bars, commonly denoted by \(| \dots |\). Absolute value represents the distance a number is from zero on a number line, irrespective of direction. For example, \(| -3 | = 3\) and \(| 3 | = 3\).

When solving inequalities with fractions and dealing with absolute value equations, you treat the material inside the absolute value bars separately. This often involves setting up cases for both the positive and negative scenarios so as to handle all possible values of the variable that make the equation true. These cases are essential because absolute value equations can produce more than one potential solution.

inequality restrictions

Inequality restrictions are essential in solving inequalities involving fractions. They help identify values of the variable that might make the denominator zero, which would make the fraction undefined.

For the inequality given as the example \( \frac{x-2}{x-4} > \frac{x+2}{x} \), we identify two key restrictions: \( x eq 4 \) and \( x eq 0 \). These values would make the fractions involved undefined and need to be excluded from the solution set.

In general, always look for points where the denominator is zero. Add these restrictions to the solution to ensure that you only consider valid values for the inequality.

common denominators

Finding common denominators is a crucial step in solving inequalities with fractions. This technique simplifies equations and inequalities, making them easier to solve.

In our exercise, the common denominator for \( \frac{x-2}{x-4} > \frac{x+2}{x} \) is \(x(x-4)\). By multiplying both sides of the inequality by this common denominator, we are able to eliminate the fractions:

• Multiply both sides: \( (x)(x-4) \cdot \frac{x-2}{x-4} > (x)(x-4) \cdot \frac{x+2}{x} \)
• Simplify: you get \( x(x-2) \) on the left and \( (x-4)(x+2) \) on the right

This step converts a complex fraction inequality into a simpler polynomial inequality: \( x(x-2) > (x-4)(x+2)\).

Remember: After simplifying the inequality by eliminating fractions, always consider the restrictions you previously identified to get the complete solution set.