Problem 43 - Algebra 2

Question

REVIEW What is the solution set of the inequality? $$6-|x+7| \leq-2$$ $$ \begin{array}{l}{\mathbf{F}-15 \leq x+\leq 1} \\ {\mathbf{G}-1 \leq x \leq 3} \\\ {\mathbf{H} x \leq-1 \text { or } x \geq 3} \\ {\mathbf{J} \quad x \leq-15 \text { or } x \geq 1}\end{array} $$

Step-by-Step Solution

Answer

The solution set is \(x \leq -15\) or \(x \geq 1\), option J.

Step 1: Isolate the Absolute Value Expression

Start by isolating the absolute value expression in the given inequality: \(6 - |x + 7| \leq -2\). To do this, subtract 6 from both sides: \[-|x+7| \leq -8\].Next, multiply the entire inequality by -1, which reverses the inequality sign: \[|x+7| \geq 8\].

Step 2: Solve the Compound Inequality

The inequality \(|x+7| \geq 8\) implies two potential cases: 1. \(x+7 \geq 8\)2. \(x+7 \leq -8\).For the first case: Solve \(x+7 \geq 8\)Subtract 7 from both sides: \(x \geq 1\).For the second case: Solve \(x+7 \leq -8\)Subtract 7 from both sides: \(x \leq -15\).

Step 3: Write the Solution Set

The two potential cases from the absolute value result in the solution set: \(x \geq 1\) or \(x \leq -15\). This indicates that any \(x\) less than or equal to -15, or \(x\) greater than or equal to 1, satisfies the original inequality.

Key Concepts

Compound Inequalities

Compound inequalities involve two separate inequalities that are combined into one statement. In this exercise, we encounter the compound inequality that arises from solving an absolute value inequality. Absolute value inequalities with the format \(|x| \geq a\) split into two separate inequalities:

• One inequality where the expression inside the absolute value is greater than or equal to the positive value: \(x \ge a\).
• Another inequality where the expression inside the absolute value is less than or equal to the negative value: \(x \le -a\).

This results in a solution set that can often look like a compound inequality involving "or" statements.
For the inequality \(|x+7| \ge 8\), solving each condition separately gives:

• Condition 1: \(x+7 \ge 8\), simplified to \(x \ge 1\).
• Condition 2: \(x+7 \le -8\), simplified to \(x \le -15\).
Thus, the solution includes all numbers where \(x \le -15\) or \(x \ge 1\), demonstrating how compound inequalities broaden the range of possible solutions.

Solution Set

The solution set of an inequality is the collection of all possible values that will satisfy the given conditions. In this scenario, the solution set includes all real numbers that satisfy either part of the compound inequality derived from the absolute value inequality.
Once the compound inequality is represented as simple statements "or" because it combines more than one condition:

• \(x \le -15\)
• \(x \ge 1\)

The union of these intervals forms the entire solution set.
This means any number less than or equal to \(-15\) or greater than or equal to \(1\) is within the solution set. This type of interval is often expressed in conjunction with inequalities and illustrates boundaries where all values satisfy the equation, forming what we call the solution set.

Algebraic Expressions

Algebraic expressions are mathematical statements involving numbers, variables, and operations. They form the basis of understanding and solving inequalities. In the provided exercise, the expression \(x+7\) nestled within absolute value signs \(|x+7|\) is an algebraic expression we need to manipulate to solve the inequality.
Isolating such expressions is crucial. Doing so often involves applying inverse operations, such as subtraction or addition, multiplication or division, always adhering to the principle of maintaining equality by treating both sides of the equation identically.

• Subtracting 6 from \(6 - |x+7| \leq -2\) isolates the absolute value: \(-|x+7| \leq -8\).
• Multiplying by \(-1\) flips the inequality sign: \(|x+7| \geq 8\).

Understanding algebraic expressions and how to manipulate them is fundamental in solving more complex mathematical problems, including those involving absolute value inequalities, as it permits us to reach solutions systematically and logically.