Solving Absolute Value Inequalities (College Algebra)

Q1. Solve the inequality for $x$: $\left| \frac{5x + 5}{2} \right| \geq 5$

Background

Topic: Absolute Value Inequalities

This question tests your understanding of how to solve inequalities involving absolute values, specifically those of the form $|X| \geq a$, where $a$ is a positive number. You need to know how to split the absolute value inequality into two separate inequalities and solve each one.

Key Terms and Formulas

• Absolute Value Inequality: For $|X| \geq a$ (where $a > 0$), the solution is $X \leq -a$ or $X \geq a$.

• Interval Notation: Used to express the solution set of inequalities.

Step-by-Step Guidance

1. Recognize the form: The inequality $\left| \frac{5x + 5}{2} \right| \geq 5$ fits the pattern $|X| \geq a$ with $X = \frac{5x + 5}{2}$ and $a = 5$.

2. Apply the property: Rewrite the inequality as two separate inequalities:

   $\frac{5x + 5}{2} \leq -5$ or $\frac{5x + 5}{2} \geq 5$

3. Clear the fractions by multiplying both sides of each inequality by 2:

   $5x + 5 \leq -10$ or $5x + 5 \geq 10$

4. Solve each inequality for $x$:

   • For the first inequality: Subtract 5 from both sides to isolate the $5x$ term.

   • For the second inequality: Do the same—subtract 5 from both sides.

5. Divide both sides of each resulting inequality by 5 to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $(-\infty, -3] \cup [1, \infty)$

After solving both inequalities, you find $x \leq -3$ or $x \geq 1$. In interval notation, this is $(-\infty, -3] \cup [1, \infty)$.

This represents all $x$ values that make the original absolute value inequality true.