Question

Solve the inequalities. $$ |x-1|<2|x-3| $$

Short answer

The solution is \((1, 7/3) \cup (5, \infty)\).

Step-by-step solution

Understand the Absolute Value Definition

The inequality \(|x - 1| < 2|x - 3|\) involves absolute values. Absolute values represent the distance of a number from zero, meaning we take the positive value of any expression inside the absolute value. To solve the inequality, we need to consider the signs of the expressions inside the absolute values and solve accordingly.

Set Up Cases Based on Critical Points

The critical points are where each absolute value expression equals zero: \(x = 1\) and \(x = 3\). Use these points to divide the number line into intervals: \((-\infty, 1)\), \((1, 3)\), and \((3, \infty)\). Consider each interval separately and remove the absolute values by considering the sign of each expression.

Solve for Case 1: \(x < 1\)

In this interval, both \(x - 1\) and \(x - 3\) are negative, so the inequality becomes \(-(x - 1) < -2(x - 3)\). This simplifies to \((-x + 1) < (-2x + 6)\). Rearranging gives \(x > 5\), which has no solutions in this interval since \(5 > 1\).

Solve for Case 2: \(1 \, < \, x \, < \, 3\)

Here, \(x - 1 > 0\) and \(x - 3 < 0\). The inequality transforms to \((x - 1) < -2(x - 3)\). This simplifies to \((x - 1) < (-2x + 6)\), leading to \(3x < 7\) or \(x < 7/3\). Since \(7/3 \approx 2.33\), and it is within \(1 < x < 3\), the solution for this case is \(1 < x < 7/3\).

Solve for Case 3: \(x > 3\)

In this interval, both \(x - 1 > 0\) and \(x - 3 > 0\), so the inequality becomes \((x - 1) < 2(x - 3)\). Simplifying gives \(x - 1 < 2x - 6\). Rearranging terms, we get \(x > 5\), which is the solution in this interval as it satisfies \(x > 3\).

Combine Solutions from All Cases

The solutions from the cases are \(x < 7/3\) for \(1 < x < 3\) and \(x > 5\). Therefore, the complete solution is the union of these intervals: \((1, 7/3) \cup (5, \infty)\).

Key concepts

Absolute Value Inequalities

Absolute value inequalities involve expressions where the variable is inside an absolute value sign. Absolute value, denoted by vertical bars like \(|x|\), represents the distance of that number from zero on a number line. It means you always get a non-negative result, no matter if the number inside is positive or negative.
Consider an inequality like \(|x-1| < 2|x-3|\). To solve this, recognize that the absolute values create a situation where we need to consider different scenarios based on the sign of the expressions inside the absolute values.
This results in multiple separate inequalities to solve. Notably, the critical points determined by setting each expression inside the absolute value to zero (here, \(x=1\) and \(x=3\)) will help define specific intervals for evaluation.

Key Steps to Solve:

• Identify the critical points where each expression within an absolute value equals zero.
• Split into cases based on intervals derived from critical points.
• Consider the sign change of the expressions as you solve within each interval.

Interval Notation

Interval notation is a compact way to describe sets of numbers, specifically solutions to inequalities. It's particularly useful in calculus for expressing where functions or inequalities hold true.
In our solution, the final answer is \( (1, 7/3) \cup (5, \infty) \). This notation describes two separate intervals where solutions to the inequality exist:

• \( (1, 7/3) \) — Numbers between 1 and 7/3 but not including 1 or 7/3.
• \( (5, \infty) \) — Numbers greater than 5 extending infinitely.

Brackets (") and parentheses (") are important here; parentheses indicate that endpoints are not included, whereas brackets would indicate inclusion.
Using interval notation efficiently communicates where solutions lie without listing every number within those intervals.

Critical Points

Critical points are crucial when dealing with inequalities involving absolute values. These are the points where the expressions inside the absolute values are equal to zero.
In the inequality \(|x - 1| < 2|x - 3|\), critical points are found by solving \(x - 1 = 0\) and \(x - 3 = 0\). This gives us \(x = 1\) and \(x = 3\).
These critical points help divide the number line into intervals. Each interval represents a different scenario for how the signs of the expressions inside the absolute values might change, which directly affects how the inequality is solved.
For example:

• Interval \( (-\infty, 1) \) considers when \(x < 1\).
• Interval \( (1, 3) \) considers when 1 < x < 3.
• Interval \( (3, \infty) \) considers when x > 3.

Critical points give structure to our approach, allowing us to handle each scenario separately and precisely.

Inequality Solving Methods

Solving inequalities, particularly those involving absolute values, involves a few key steps and strategies. Understanding these methodologies is crucial for efficiently finding solutions.
The main processes include:

• Identify and verify critical points. This sets the stage for solving by dividing the problem into manageable sections based on these points.
• Consider each interval separately. By looking at each interval formed by critical points, you determine how to handle the inequality without absolute values. This often involves reversing inequality signs based on positive or negative values.
• Simplify each form of the inequality within the interval. For each case, simplify and solve the inequality, maintaining careful attention to the direction of inequality signs, especially after multiplying or dividing by a negative number.
• Combine solutions. After solving for each scenario individually, compile the solutions using union notation to represent all values that satisfy the original inequality.

Using these methods helps in logically unraveling complex inequalities, ensuring that no solution is overlooked. Each method applied serves to break down and clarify, resulting in a coherent and comprehensive solution.