Problem 40 - Algebra and Trigonometry

Question

Solve the inequality. Express the answer using interval notation. $$ 2\left|\frac{1}{2} x+3\right|+3 \leq 51 $$

Step-by-Step Solution

Answer

\([-54, 42]\)

Step 1: Isolate the Absolute Value

First, isolate the absolute value expression \(2\left|\frac{1}{2} x+3\right|+3 \leq 51\) by subtracting 3 from both sides of the inequality.\[2\left|\frac{1}{2} x+3\right| \leq 48\]

Step 2: Solve for the Absolute Value Expression

Divide both sides by 2 to solve for \(\left|\frac{1}{2} x+3\right|\).\[\left|\frac{1}{2} x+3\right| \leq 24\]

Step 3: Split into Two Inequalities

The expression \(\left|A\right| \leq B\) splits into two inequalities: \(A \leq B\) and \(A \geq -B\). Apply this to \(\left|\frac{1}{2} x+3\right|\):1. \(\frac{1}{2} x + 3 \leq 24\)2. \(\frac{1}{2} x + 3 \geq -24\)

Step 4: Solve the First Inequality

Begin with the inequality \(\frac{1}{2} x + 3 \leq 24\). Subtract 3 from both sides:\[\frac{1}{2} x \leq 21\]Then multiply both sides by 2 to solve for \(x\):\[x \leq 42\]

Step 5: Solve the Second Inequality

Now solve \(\frac{1}{2} x + 3 \geq -24\). Subtract 3 from both sides:\[\frac{1}{2} x \geq -27\]Then multiply both sides by 2 to solve for \(x\):\[x \geq -54\]

Step 6: Combine the Solutions

The solutions from Steps 4 and 5 are combined as \(-54 \leq x \leq 42\). In interval notation, this is:\([-54, 42]\).

Key Concepts

Absolute Value Inequalities

In mathematics, understanding absolute value inequalities is crucial. These inequalities involve expressions within absolute value symbols, such as \(\|A\| \leq B\), where A can have a range of values. The absolute value of a number represents its distance from zero on the number line. This means that the solutions can be positive or negative, as long as they satisfy the inequality.When solving absolute value inequalities, the key is to "split" the inequality into two separate inequalities. This is due to the nature of absolute value, which encompasses both positive and negative possibilities. For example, in the expression \(\|\frac{1}{2} x + 3\| \leq 24\), we derive two inequalities: \(\frac{1}{2} x + 3 \leq 24\) and \(\frac{1}{2} x + 3 \geq -24\). Thus, the solution represents all x values that make either inequality true.

Interval Notation

Interval notation is a concise way of expressing sets of numbers, typically representing the solution to inequalities on a number line. It indicates the range of values included in or excluded from the solution. For example, \(-54 \leq x \leq 42\) leads to the interval notation \([-54, 42]\).In interval notation:

• Square brackets \([ ]\) indicate that the endpoints are included in the interval (closed interval).
• Parentheses \( ( ) \) show that the endpoints are not part of the interval (open interval).

This format efficiently communicates the set of numbers that satisfy an inequality. Knowing how to read and write interval notation is essential for expressing solutions in a mathematical context.

Step-by-step Algebra Solutions

Approaching algebra problems with a step-by-step method is vital for clarity and accuracy. This systematic process involves solving equations or inequalities incrementally. Let's break down the approach used in the given solution.

1. Isolate the Absolute Value

To start, isolating the absolute value expression helps simplify further steps. In the inequality \(2\left|\frac{1}{2} x+3\right|+3 \leq 51\), subtract 3 from both sides, resulting in \(2\left|\frac{1}{2} x+3\right| \leq 48\).

2. Solve for the Absolute Value

Divide both sides by 2, yielding \(\left|\frac{1}{2} x+3\right| \leq 24\). Splitting this absolute value leads to two inequalities.

3. Convert to Two Linear Inequalities

By separating into \(\frac{1}{2} x + 3 \leq 24\) and \(\frac{1}{2} x + 3 \geq -24\), you handle the positive and negative scenarios.

4. Solve Each Inequality

Solve each individually by simplifying further. Subtract 3, then multiply by 2 for each part, yielding \(x \leq 42\) and \(x \geq -54\).Through these guided steps, each part of the process can be clearly followed, ensuring a full understanding of solving inequalities.