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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.

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.

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.