Problem 44 - Algebra and Trigonometry

Question

Solve the inequality. Express the answer using interval notation. $$ \frac{1}{|2 x-3|} \leq 5 $$

Step-by-Step Solution

Answer

\((-\infty, \frac{7}{5}] \cup [\frac{8}{5}, \infty)\)

Step 1: Understand the Given Inequality

The inequality \( \frac{1}{|2x - 3|} \leq 5 \) involves the expression within absolute value and dealing with a reciprocal function. We need to isolate the absolute value expression to solve further.

Step 2: Rewrite the Inequality

Multiply both sides of the inequality by \(|2x - 3|\) (assuming \(|2x - 3| eq 0\)), obtaining \(1 \leq 5|2x - 3|\).

Step 3: Simplify the Inequality

To simplify, divide both sides by 5, giving \(\frac{1}{5} \leq |2x - 3|\). This implies that \(|2x - 3| \geq \frac{1}{5}\).

Step 4: Solve the Absolute Value Inequality

The inequality \(|2x - 3| \geq \frac{1}{5}\) translates to two inequalities: 1. \(2x - 3 \geq \frac{1}{5}\)2. \(2x - 3 \leq -\frac{1}{5}\)

Step 5: Solve the First Inequality

Solving \(2x - 3 \geq \frac{1}{5}\):- Add 3 to both sides: \(2x \geq \frac{1}{5} + 3\)- Simplify the addition: \(2x \geq \frac{16}{5}\)- Divide by 2: \(x \geq \frac{8}{5}\).

Step 6: Solve the Second Inequality

Solving \(2x - 3 \leq -\frac{1}{5}\):- Add 3 to both sides: \(2x \leq -\frac{1}{5} + 3\)- Simplify the addition: \(2x \leq \frac{14}{5}\)- Divide by 2: \(x \leq \frac{7}{5}\).

Step 7: Write the Solution in Interval Notation

Combine the solutions from Step 5 and Step 6. Since the inequality represents \(|2x - 3|\) being greater than or equal to \(\frac{1}{5}\), the solutions are separate intervals:- When \(2x - 3 \geq \frac{1}{5}\): \(x \geq \frac{8}{5}\)- When \(2x - 3 \leq -\frac{1}{5}\): \(x \leq \frac{7}{5}\)Thus, the solution in interval notation is: \((-\infty, \frac{7}{5}] \cup [\frac{8}{5}, \infty)\).

Key Concepts

Understanding Absolute Value Inequalities

Absolute value inequalities like \(|2x - 3| \geq \frac{1}{5}\) can seem tricky, but they're just a way of expressing distance on the number line.
The absolute value symbol \(|\cdot|\) dictates how far a number is from zero, so when we deal with inequalities, we focus on the notion of distance from a point.
When solving an absolute value inequality, break it down into two separate inequalities:

• For \(|a| \geq b\), you translate to both \(a \geq b\) and \(a \leq -b\).

This means you're looking for values either greater than a certain number or less than its negative counterpart.
In our exercise, this translates \(|2x - 3| \geq \frac{1}{5}\) into two parts: \(2x - 3 \geq \frac{1}{5}\) and \(2x - 3 \leq -\frac{1}{5}\).
Solving each gives us intervals that capture all solutions satisfying the inequality.

Expressing Solutions in Interval Notation

Interval notation is a concise way to express solution sets for inequalities.
It uses brackets \([ \text{and} ]\) for inclusive bounds and parentheses \(( \text{and} )\) for exclusive bounds.
In solutions like these, consider:

• \([a, b]\) means all numbers from \(a\) to \(b\) are included.
• \((a, b)\) means numbers between \(a\) and \(b\) but not including \(a\) or \(b\).
• To represent infinity, always use \((\infty)\) or \((-\infty)\), as infinity cannot be a specific endpoint.

For the exercise, the solutions \(( -\infty, \frac{7}{5} ] \cup [ \frac{8}{5}, \infty )\) show how multiple intervals are merged together with a union symbol \(\cup\), capturing two distinct ranges of \(x\) values.
This offers a neat way of visualizing and interpreting solutions to inequalities.

Demystifying Reciprocal Functions in Inequalities

Reciprocal functions, such as \(\frac{1}{|2x - 3|}\), involve flipping a number to get its "reciprocal".
For instance, the reciprocal of \(x\) is \(\frac{1}{x}\).
In inequalities, especially with reciprocals, remember these points:

• Reciprocals switch the direction of the inequality when multiplied across, except in absolute terms.
• The inequality needs extra care to ensure the expression within isn't zero, as division by zero is undefined.

In our exercise, we handle this by allowing \(|2x - 3|\) to be non-zero, simplifying \(1 \leq 5|2x - 3|\).
Solving zones where \(|2x - 3| = 0\) ensures complete accuracy by avoiding undefined operations.
Understanding this makes handling reciprocals intuitive in the realm of inequalities.