Problem 47 - Algebra and Trigonometry

Question

\(23-48\) Solve the inequality. Express the answer using interval notation. $$ \frac{1}{|x+7|}>2 $$

Step-by-Step Solution

Answer

The solution in interval notation is \((-\frac{15}{2}, -\frac{13}{2})\).

Step 1: Write the Inequality in a Solvable Form

Start with the inequality given: \( \frac{1}{|x+7|} > 2 \). We want to eliminate the fraction by taking the reciprocal on both sides. This turns the inequality into \( |x+7| < \frac{1}{2} \).

Step 2: Break Down the Absolute Value Inequality

An inequality involving absolute value \( |A| < B \) implies two inequalities: \( A < B \) and \( A > -B \). For our inequality \( |x+7| < \frac{1}{2} \), we write: \( x+7 < \frac{1}{2} \) and \( x+7 > -\frac{1}{2} \).

Step 3: Solve Each Inequality Separately

Let's solve the first inequality: \( x+7 < \frac{1}{2} \). Subtract 7 from both sides to get \( x < \frac{1}{2} - 7 \). Simplifying gives \( x < -\frac{13}{2} \). Now, for the second inequality: \( x+7 > -\frac{1}{2} \). Subtract 7 from both sides to get \( x > -\frac{1}{2} - 7 \). Simplifying gives \( x > -\frac{15}{2} \).

Step 4: Combine the Solutions

The solutions from Step 3 are \( x < -\frac{13}{2} \) and \( x > -\frac{15}{2} \). Combine these to find the interval that satisfies both inequalities: \(-\frac{15}{2} < x < -\frac{13}{2}\).

Step 5: Express the Answer in Interval Notation

Convert the inequality \(-\frac{15}{2} < x < -\frac{13}{2}\) into interval notation. This means the solution is the open interval \((-\frac{15}{2}, -\frac{13}{2})\).

Key Concepts

Absolute Value Inequalities

When you see the term absolute value inequality, it's all about understanding how far numbers are from zero on a number line. Absolute value is denoted by two vertical bars, like this: \(|x|\). The absolute value of a number is always positive or zero. It's important because it measures the "distance" regardless of direction.
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For absolute value inequalities, such as \(|A| < B\), you're essentially setting up two conditions:

• \(A < B\)
• \(A > -B\)

This tells you that the value of \(A\) is squeezed between \(-B\) and \(B\). In the exercise, \(|x + 7| < \frac{1}{2}\) breaks down into two simple inequalities:

• \(x + 7 < \frac{1}{2}\)
• \(x + 7 > -\frac{1}{2}\)

These inequalities are then simplified to find the range of \(x\). Understanding that absolute value inequalities give you two paths to solve can make these problems less intimidating, because you're just solving two straightforward inequalities.

Interval Notation

Interval notation is a way to express solution sets involving inequalities. It helps indicate where a solution starts and ends on the number line without explicitly listing every number. This notation uses parentheses or brackets to show whether endpoints are included or not.

• A parenthesis \((\) or \()\) means the endpoint is not included.
• A bracket \([\) or \()]\) means the endpoint is included.

In the context of the exercise, we combined the two inequalities into one to get \(-\frac{15}{2} < x < -\frac{13}{2}\). These solutions are expressed as an open interval, which means neither endpoint is included. Therefore, the interval notation becomes \((-\frac{15}{2}, -\frac{13}{2})\).
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Interval notation is a great shorthand that makes the solution quick to write and easy to understand, especially when dealing with inequalities. Instead of using phrases like "greater than" and "less than," interval notation is precise and mathematical.

Fractional Inequalities

Fractional inequalities involve expressions that contain fractions. Solving these requires careful manipulation to avoid violations of inequality properties, especially when dealing with negative numbers. To solve them, we often need to remove the fraction by multiplying through by the denominator or taking reciprocal steps, just like we did in the exercise.

• When you have \(\frac{1}{|x+7|} > 2\), you start by taking the reciprocal to simplify the inequality.
• Turning the inequality into \(|x+7| < \frac{1}{2}\) lets you tackle it as an absolute value inequality.

Keep two things in mind: when you multiply or divide by a negative, you must reverse the inequality sign. Also, remember to check that solutions don't make any original denominators zero, as that would make the expression undefined. In our example, since the value of \(x\) doesn't make \(|x+7| = 0\), we safely proceed with our solution.