Problem 70 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ 2<|2 x-1|<3 $$

Step-by-Step Solution

Answer

The solution is \((-1, -\frac{1}{2}) \cup (\frac{3}{2}, 2)\).

Step 1: Understanding Absolute Inequality

The inequality \(2 < |2x - 1| < 3\) involves an absolute value expression between two limits. Our task is to solve for \(x\) such that the absolute value of \(2x - 1\) lies strictly between 2 and 3.

Step 2: Splitting the Inequality

Since the absolute value expression is between two numbers, we have two separate compound inequalities to solve: \(2 < |2x - 1|\) and \(|2x - 1| < 3\).

Step 3: Solving \(2 < |2x - 1|\)

The absolute inequality \(2 < |2x - 1|\) splits into two cases:1. \(2x - 1 > 2\)2. \(2x - 1 < -2\).Solving the first case: \(2x - 1 > 2\) leads to \(2x > 3\), thus \(x > \frac{3}{2}\).Solving the second case:\(2x - 1 < -2\) leads to \(2x < -1\), thus \(x < -\frac{1}{2}\).

Step 4: Solving \(|2x - 1| < 3\)

The inequality \(|2x - 1| < 3\) also splits into two cases:1. \(2x - 1 < 3\)2. \(2x - 1 > -3\).Solving the first case:\(2x - 1 < 3\) leads to \(2x < 4\), thus \(x < 2\).Solving the second case:\(2x - 1 > -3\) leads to \(2x > -2\), thus \(x > -1\).

Step 5: Combining Solutions

We have the following ranges:1. From \(2 < |2x - 1|\): \(x > \frac{3}{2}\) or \(x < -\frac{1}{2}\).2. From \(|2x - 1| < 3\): \(-1 < x < 2\).Combining these conditions, we find that:- \(x\) must satisfy both sets of inequalities simultaneously.

Step 6: Determining the Intersection

Intersect the two solution sets:- The interval for \(x > \frac{3}{2}\) with \(-1 < x < 2\) gives \(\frac{3}{2} < x < 2\).- The interval for \(x < -\frac{1}{2}\) with \(-1 < x < 2\) gives \(-1 < x < -\frac{1}{2}\).Thus, the solution in interval notation is the union of these ranges: \((-1, -\frac{1}{2}) \cup (\frac{3}{2}, 2)\).

Key Concepts

Absolute Inequalities

An absolute inequality is an inequality that involves an absolute value expression. If you see a problem like \(2 < |2x - 1| < 3\), you're dealing with absolute inequalities.

When solving absolute inequalities, the absolute value splits the inequality into two different cases because the absolute value of a number is its distance from zero—regardless of direction. This gives us two possible scenarios:

• Case 1: The expression inside the absolute value is positive or zero.
• Case 2: The expression inside the absolute value is negative.

To solve \(2 < |2x - 1|\), for example, we consider both \(2x - 1 > 2\) and \(2x - 1 < -2\). Solving these will give you an idea of the values \(x\) can take while adhering to the inequality. The key point is to always split the inequality and solve each separately.

Interval Notation

Interval notation is a shorthand method to express sets of numbers, specifically the sets that contain the solutions of inequalities. If you've solved an inequality and found that \(-1 < x < -\frac{1}{2}\) or \(\frac{3}{2} < x < 2\), you can express these ranges in interval notation as \((-1, -\frac{1}{2})\) and \((\frac{3}{2}, 2)\) respectively.

In interval notation, parantheses \(()\) are used to indicate that an endpoint is not included in the interval, known as an "open interval." Meanwhile, brackets \([]\) would indicate that the number is included, known as a "closed interval."

• Open Interval: \((a, b)\) means \(a < x < b\).
• Closed Interval: \([a, b]\) means \(a \leq x \leq b\).
• Half-Open (or Half-Closed) Intervals: \([a, b)\) or \((a, b]\).

Interval notation provides a concise way of describing complex solution sets.

Compound Inequalities

Compound inequalities are expressions involving two distinct inequalities joined together, often by the words 'and' or 'or'.

For instance, consider solving the inequality \(2 < |2x - 1| < 3\). This can be viewed as a compound inequality as we determined two separate inequalities: \(2 < |2x - 1|\) and \(|2x - 1| < 3\).

When solving compound inequalities, themselves create intersections or unions of possible solutions:

• "And" compound inequalities require both conditions to be true simultaneously. Their solution is the intersection or overlap of the sets.
• "Or" compound inequalities require at least one condition to be true. This reflects in the union of the sets.

When you have reached solutions for each part of a compound inequality, like combining \(x > \frac{3}{2}\) with \(-1 < x < 2\), only values that satisfy both are considered solutions, using intersections to find valid solutions. If dealing with "or" compound inequalities, use the union of sets such as \((-1, -\frac{1}{2}) \cup (\frac{3}{2}, 2)\). Understanding intersections and unions helps to systematically determine the final solution.