Problem 50 - Precalculus Student Solutions Manual 5th

Question

Solve each equation or inequality. $$\left|2 x+\frac{1}{3}\right|+1<4$$

Step-by-Step Solution

Answer

\( -\frac{5}{3} < x < \frac{4}{3} \)

Step 1: Isolate the Absolute Value Term

To start, isolate the absolute value term by subtracting 1 from both sides of the equation: \[ \left|2x + \frac{1}{3}\right| + 1 < 4 \Rightarrow \left|2x + \frac{1}{3}\right| < 3 \]

Step 2: Set Up Two Inequalities

The expression \( \left|2x + \frac{1}{3}\right| < 3 \) can be broken into two inequalities: \[ 2x + \frac{1}{3} < 3 \] and \[ 2x + \frac{1}{3} > -3 \]

Step 3: Solve the First Inequality

First, solve the inequality \( 2x + \frac{1}{3} < 3 \):1. Subtract \(\frac{1}{3}\) from both sides: \[ 2x < 3 - \frac{1}{3} \Rightarrow 2x < \frac{9}{3} - \frac{1}{3} \Rightarrow 2x < \frac{8}{3} \]2. Divide by 2: \[ x < \frac{8}{3} \div 2 \Rightarrow x < \frac{4}{3} \]

Step 4: Solve the Second Inequality

Now, solve the inequality \( 2x + \frac{1}{3} > -3 \):1. Subtract \(\frac{1}{3}\) from both sides: \[ 2x > -3 - \frac{1}{3} \Rightarrow 2x > -\frac{9}{3} - \frac{1}{3} \Rightarrow 2x > -\frac{10}{3} \]2. Divide by 2: \[ x > -\frac{10}{3} \div 2 \Rightarrow x > -\frac{5}{3} \]

Step 5: Combine the Inequalities

Combine the solutions of the two inequalities: \[ -\frac{5}{3} < x < \frac{4}{3} \]

Key Concepts

Solving Inequalities

Inequalities show the relationship between expressions that are not necessarily equal. They come in forms such as <, >, ≤, and ≥. To solve inequalities, follow these basic steps:

1. **Isolate the variable**: Just like solving equations, get the variable by itself on one side of the inequality. Use operations such as addition, subtraction, multiplication, and division.
2. **Inverse operations**: Apply inverse operations to both sides of the inequality to maintain balance. For example, if the inequality is multiplied by a number, you divide by that number.
3. **Flip the inequality sign**: If you multiply or divide both sides of an inequality by a negative number, remember to flip the inequality sign.

Once you have isolated the variable, interpret the solution. For example, if the solution is \(x < 5\), it means that x can be any value less than 5. Keep practicing, and solving inequalities will become second nature!

Absolute Value

Absolute value represents the distance of a number from zero on a number line, always given as a non-negative value. It's symbolized by two vertical bars, like this: \(|x|\).

Here are the key points about absolute value:

* **Non-negativity**: The result is always zero or positive. For example, \(|3| = 3\) and \(|-3| = 3\).
* **Two cases**: To solve equations or inequalities involving absolute values, you need to consider both the positive and negative cases of the expressions inside. For instance, if \( |x| < 4 \), this translates to \( x < 4 \) and \(-x < 4\) (which simplifies to \( x > -4 \)).

In the context of inequalities, solving for the variable inside an absolute value often involves setting up and solving two separate inequalities. This allows you to capture the range of values that satisfy the condition.

Interval Notation

Interval notation is a way of writing subsets of the real number line. It uses brackets to show closed intervals (where endpoints are included) and parentheses for open intervals (where endpoints are not included). Here’s a simple breakdown:

* **Open interval (a, b)**: All numbers between a and b, but not including a and b. Example: \( (1, 3) \) represents all numbers greater than 1 and less than 3.
* **Closed interval [a, b]**: All numbers from a to b, including both a and b. Example: \( [1, 3] \) includes 1, 3, and all numbers in between.
* **Half-open intervals**: Either endpoint may be included. Example: \( [1, 3) \) includes 1 but not 3.

In inequalities, once you find the solution set, you can easily translate this into interval notation. For instance, if you solve an inequality and find that the solution is \(-2 < x < 5\), the interval notation would be \((-2, 5)\).