Problem 41 - Beginning and Intermediate Algebra

Question

Solve each inequality. Graph the solution set and write the answer in interval notation. $$|2 m-1|+4>5$$

Step-by-Step Solution

Answer

Short Answer: The inequality \(|2m-1|+4>5\) has no solution. The solution set is empty, which we represent graphically by not highlighting any region and in interval notation as \(∅\).

Step 1: Remove the absolute value symbol

To remove the absolute value symbol, we rewrite the inequality as a system of two inequalities, one for the positive case and one for the negative case: Positive case: \(2m-1 + 4 > 5\) Negative case: \(-(2m-1) + 4 > 5\)

Step 2: Solve the positive case inequality

Now, we will solve the positive case inequality: \(2m - 1 + 4 > 5\) Combine the constants: \(2m + 3 > 5\) Subtract 3 from both sides: \(2m > 2\) Divide by 2: \(m > 1\)

Step 3: Solve the negative case inequality

Now, we will solve the negative case inequality: \(-(2m - 1) + 4 > 5\) Distribute the negative sign: \(-2m + 1 + 4 > 5\) Combine the constants: \(-2m + 5 > 5\) Subtract 5 from both sides: \(-2m > 0\) Divide by -2 (remember to flip the inequality sign when multiplying or dividing by a negative number): \(m < 0\)

Step 4: Combine the solutions

Now, we will combine the solutions from Step 2 and Step 3. Since there is no overlap between the two solution sets, meaning no single value of m will satisfy both inequalities simultaneously, it means the absolute value inequality has no solution. In this case, the solution set is empty.

Step 5: Graph the solution set

Since the solution set is empty, there are no values of m that satisfy the inequality. In the graph, we will not have any points or interval to highlight. An empty graph with no highlighted region represents this situation.

Step 6: Write the answer in interval notation

Since the solution set is empty, there are no intervals to represent the solutions. We can denote the empty solution set using the empty set symbol, which is ∅ in interval notation. Thus, the final answer is given by: Solution: \(∅\)

Key Concepts

Absolute Value Inequalities

Absolute value inequalities involve finding the range of values for a variable that keeps the inequality true when inside absolute value bars. The crucial part is understanding how to break them down. For an inequality like \(|ax + b| > c\), we split it into two separate cases:

• Positive case: \(ax + b > c\)
• Negative case: \(-(ax + b) > c\)

This means we're considering both directions from zero. Herein lies the trick: absolute value measures distance, so you explore scenarios on both sides. Solving each separately before combining ensures you cover all possible values.

System of Inequalities

When dealing with absolute value inequalities, we often end up with a system of inequalities. Each inequality corresponds to one scenario of the absolute value expression.
For instance, we solve for \(m\) as follows:

• The positive case: \(2m + 3 > 5\)
• The negative case: \(-2m + 5 > 5\)

Each inequality gives a range of possible values. By solving each, we gather potential solutions to verify which align with the problem's constraints. This approach ensures comprehensive coverage of all possibilities.

Interval Notation

Interval notation is a concise way to express the set of solutions for inequalities. When writing an interval:

• A parenthesis \(()\) indicates that a number is not included (open interval).
• A bracket \([]\) shows that a number is included (closed interval).

For example, the solution \(m > 1\) translates to \((1, \infty)\) in interval notation. Using this notation, we can quickly communicate the range of values without the need for further explanation. It's a powerful tool for clarity in mathematical solutions.

Empty Set Solution

Sometimes, solving an inequality reveals that no values satisfy all conditions. In these cases, the solution set is empty.
For example, if none of the solutions from individual cases overlap, there are no common values. This is expressed in interval notation as the empty set, denoted by \(\emptyset\).
In graph terms, this translates to no regions being shaded. Understanding this concept is crucial because it signifies that while there may be potential solutions in different scenarios, none fit comprehensively.