If x is positive real number, find the range of values for which $\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$

Analysis

We are given the inequality $\frac{1}{3x} + \frac{1}{2} > \frac{1}{4x}$ where $x$ is a positive real number. Our goal is to isolate $x$ to find the range of values that satisfy this condition.

Concept

To solve inequalities involving fractions, we need to:

1. Find a common denominator to combine the terms.
2. Simplify the inequality.
3. Apply the properties of inequalities (remembering that multiplying or dividing by a negative number flips the inequality sign).

Step-by-Step Solution

Step 1: Move all terms to one side. $\frac{1}{3x} - \frac{1}{4x} + \frac{1}{2} > 0$

Step 2: Find a common denominator for the $x$ terms. The common denominator for $3x$ and $4x$ is $12x$. $\frac{4 - 3}{12x} + \frac{1}{2} > 0$ $\frac{1}{12x} + \frac{1}{2} > 0$

Step 3: Combine the fractions. To add $\frac{1}{12x}$ and $\frac{1}{2}$, the common denominator is $12x$: $\frac{1 + 6x}{12x} > 0$

Step 4: Analyze the inequality. We are given that $x$ is a positive real number ($x > 0$). If $x > 0$, then $12x$ is definitely positive. For the fraction $\frac{1 + 6x}{12x}$ to be greater than zero (positive), the numerator $(1 + 6x)$ must also be positive because a positive divided by a positive is positive.

$1 + 6x > 0$ $6x > -1$ $x > -\frac{1}{6}$

Step 5: Consider the constraint. The problem explicitly states that $x$ is a positive real number ($x > 0$). Since our calculated range $x > -\frac{1}{6}$ includes negative numbers, we must intersect this with the given condition $x > 0$. The intersection of $x > -\frac{1}{6}$ and $x > 0$ is simply $x > 0$.

Conclusion

The inequality holds true for all positive values of $x$. Therefore, the range is $x > 0$.

Exam Tip

When dealing with inequalities involving variables in the denominator, always check if the variable is restricted (e.g., $x > 0$). If you multiply both sides by $x$ to clear the fraction, ensure you know the sign of $x$ so you know whether to flip the inequality sign!