Problem 9 - Algebra 2

Question

Solve each inequality. $$ \frac{1}{3 v}+\frac{1}{4 v}<\frac{1}{2} $$

Step-by-Step Solution

Answer

\( v \in \left( \frac{7}{6}, \infty \right) \)

Step 1: Combine the Fractions on the Left

Add the fractions \( \frac{1}{3v} \) and \( \frac{1}{4v} \) to have a single fraction on the left-hand side. To add these fractions, the common denominator is \( 12v \). Therefore: \[ \frac{1}{3v} + \frac{1}{4v} = \frac{4 + 3}{12v} = \frac{7}{12v}. \] Rewrite the inequality: \[ \frac{7}{12v} < \frac{1}{2}. \]

Step 2: Eliminate the Denominator

To eliminate the fraction, multiply both sides of the inequality by \( 12v \) (note: assume \( v eq 0 \) since division by zero is undefined). This gives: \[ 7 < \frac{1}{2} \cdot 12v. \] This simplifies to: \[ 7 < 6v. \]

Step 3: Solve for v

To isolate \( v \), divide both sides of the inequality by 6: \[ \frac{7}{6} < v. \] This means \( v \) is greater than \( \frac{7}{6} \).

Step 4: Express the Solution

Write the solution in interval notation. Since \( v > \frac{7}{6} \), the solution is \( v \in \left( \frac{7}{6}, \infty \right) \).

Key Concepts

Understanding Rational Inequalities

Rational inequalities involve expressions with fractions where you compare values using inequality symbols such as <, >, ≤, and ≥. In these inequalities, you'll often find both the numerators and denominators contain variables.
For example, the inequality \( \frac{1}{3v} + \frac{1}{4v} < \frac{1}{2} \) is a rational inequality because it includes fractions with the variable \( v \) in the denominator.
Here, solving rational inequalities requires you to approach the problem thoughtfully, often by finding a common denominator or by eliminating the fractions methodically.
It's crucial to remember that division by zero is undefined, so the variable in the denominator should not make the expression equal zero. Be mindful of this when finding a solution set.

Fraction Operations for Rational Inequalities

Performing fraction operations accurately is key to solving rational inequalities. When adding or subtracting fractions, the first step is to find a common denominator.

• In the given exercise, the fractions \( \frac{1}{3v} \) and \( \frac{1}{4v} \) are combined by using a common denominator of \( 12v \).
• Adding these fractions can be expressed as \( \frac{4}{12v} + \frac{3}{12v} = \frac{7}{12v} \).

Once the fractions are combined, you can multiply both sides of the inequality by this common denominator to eliminate the fractions, simplifying the expression. This leads to an easier inequality equation to solve, such as\( 7 < 6v \).
After clearing the fractions, the next step is to isolate the variable. This often involves simple algebraic manipulations like dividing or multiplying both sides by a constant.

Using Interval Notation in Solutions

Interval notation is a concise way to express a range of numbers, often used to represent solutions of inequalities.

• Intervals are written with parentheses \(( )\) or brackets \([ ]\). Parentheses mean that the endpoint is not included, while brackets mean the endpoint is included.

For the solution of \( \frac{7}{6} < v \), interval notation would be \( v \in \left( \frac{7}{6}, \infty \right) \). Here:

• The parentheses around \( \frac{7}{6} \) indicate that this endpoint is not part of the solution.
• \( \infty \) always uses a parenthesis since infinity is not a finite value.

Interval notation gives a clear picture of the scope of solutions, making it an essential tool for communicating inequality answers effectively in mathematics.