Problem 21 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 26 , find the solution set of the given inequality, and illustrate the solution on the real number line. $$ |x+4| \leq|2 x-6| $$

Step-by-Step Solution

Answer

The solution set is \( -4 \leq x \leq \frac{2}{3} \) and \( x \geq 10 \).

Step 1: Understand the given inequality

The inequality is given as \[ |x+4| \leq |2x-6|. \]Recognize that both sides of the inequality involve absolute values.

Step 2: Consider different cases for absolute values

Absolute value expressions can be split into different cases based on the value inside the absolute value. So, we need to consider:1. When both expressions inside the absolute values are non-negative.2. When one of them is negative.

Step 3: Case 1 - Both expressions are non-negative

If both expressions inside the absolute value are non-negative, we have: \( x + 4 \geq 0 \) and \( 2x - 6 \geq 0 \).This simplifies to \( x \geq -4 \) and \( x \geq 3 \).So, we only need to consider \( x \geq 3 \).In this range, the inequality becomes:\[ x + 4 \leq 2x - 6. \]Simplify this to:\[ 10 \leq x. \]

Step 4: Case 2 - First expression non-negative, second expression negative

In this case,\( x + 4 \geq 0 \) and \( 2x - 6 \leq 0 \).This simplifies to \( x \geq -4 \) and \( x \leq 3 \).So, we need to consider the range \( -4 \leq x \leq 3 \).in this interval, the inequality becomes:\[ x + 4 \leq -(2x - 6). \]Simplify this to: \[ x + 4 \leq -2x + 6. \]Combine like terms:\[ 3x \leq 2 \] which simplifies to: \[ x \leq \frac{2}{3}. \]So the interval is \( -4 \leq x \leq \frac{2}{3}. \)

Step 5: Case 3 - First expression negative, second expression non-negative

In this case,\( x + 4 \leq 0 \) and \( 2x - 6 \geq 0 \).This simplifies to \( x \leq -4 \) and \( x \geq 3 \).There is no value of \( x \) that satisfies both conditions simultaneously, so this case is not possible.

Step 6: Case 4 - Both expressions are negative

In this case,\( x + 4 \leq 0 \) and \( 2x - 6 \leq 0 \).This simplifies to \( x \leq -4 \) and \( x \leq 3 \).So we need to consider the range \( x \leq -4. \)In this interval, the inequality becomes:\[ -(x + 4) \leq -(2x - 6). \]Simplify this to:\[ -x - 4 \leq -2x + 6. \]Combine like terms:\[ -x + 2x \leq 6 + 4. \]So:\[ x \leq 10. \]Considering the range given for this case, we have \( x \leq -4. \)

Step 7: Combine the intervals

Combine the intervals from the valid cases, we get:\[ -4 \leq x \leq \frac{2}{3} \] and \[ x \geq 10. \]

Step 8: Illustrate the solution on the real number line

Draw a real number line and shade the regions that satisfy \( -4 \leq x \leq \frac{2}{3} \) and \( x \geq 10. \)

Key Concepts

absolute value inequality

The given problem involves an absolute value inequality: \(|x+4| \leq |2x-6|\). An absolute value represents the distance of a number from zero on the number line, regardless of direction. Thus, \( |a| = a \) if \(a \geq 0\), and \( |a| = -a \) if \(a < 0\). When solving absolute value inequalities, we consider multiple scenarios based on different possible values of the expressions inside the absolute values. This process is necessary to capture all potential solutions.

real number line

The real number line is essential in visualizing and understanding the solution sets of inequalities. The number line represents all possible values of a variable along a continuous line. Each point corresponds to a real number.
An inequality solution often specifies a range of values, which can be visualized as a shaded region on the number line. For the given problem, the solution can be broken into intervals:

• From \( -4\) to \( \frac{2}{3} \) inclusive
• From \(10\) onwards

These intervals are shaded on the number line to represent the complete set of solutions.

case analysis in inequalities

Case analysis is a critical strategy in solving absolute value inequalities. Here, we consider different scenarios based on the signs (positive or negative) of the expressions inside the absolute values. In the given problem, there are four possible cases:
Each case helps break down the problem into simpler parts:
1. Both expressions are non-negative \((x+4 \geq 0 \text{ and } 2x-6 \geq 0)\).
2. The first expression is non-negative, and the second expression is negative \((x+4 \geq 0\text{ and } 2x-6 \leq 0)\).
3. The first expression is negative, and the second expression is non-negative \((x+4 \leq 0\text{ and } 2x-6 \geq 0)\).
4. Both expressions are negative \((x+4 \leq 0 \text{ and } 2x-6 \leq 0)\).
By addressing these cases one by one, we derive ranges of \(x\) values where the original inequality holds. This method ensures comprehensive coverage of all possible solutions.