Solve the inequality.

StartAbsoluteValue 6 minus 5 x EndAbsoluteValue plus 1 greater than or equals 9

To solve the inequality \( |6 - 5x| + 1 \geq 9 \), follow these steps:

### Step 1: Isolate the absolute value expression
Subtract 1 from both sides of the inequality to isolate the absolute value:
\[
|6 - 5x| + 1 - 1 \geq 9 - 1
\]
\[
|6 - 5x| \geq 8
\]

### Step 2: Remove the absolute value by considering two cases
The absolute value inequality \( |A| \geq B \) implies two possible cases:
1. \( A \geq B \)
2. \( A \leq -B \)

Applying this to our inequality:
\[
6 - 5x \geq 8 \quad \text{or} \quad 6 - 5x \leq -8
\]

### Step 3: Solve each inequality

**Case 1: \( 6 - 5x \geq 8 \)**
\[
6 - 5x \geq 8
\]
Subtract 6 from both sides:
\[
-5x \geq 2
\]
Divide by -5, and remember to reverse the inequality sign because we are dividing by a negative number:
\[
x \leq -\frac{2}{5}
\]

**Case 2: \( 6 - 5x \leq -8 \)**
\[
6 - 5x \leq -8
\]
Subtract 6 from both sides:
\[
-5x \leq -14
\]
Divide by -5, and reverse the inequality sign:
\[
x \geq \frac{14}{5}
\]

### Step 4: Combine the results
The solution to the inequality is:
\[
x \leq -\frac{2}{5} \quad \text{or} \quad x \geq \frac{14}{5}
\]
In interval notation, this is:
\[
\left( -\infty, -\frac{2}{5} \right] \cup \left[ \frac{14}{5}, \infty \right)
\]

Would you like more details or have any questions?

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**Related Questions:**
1. How do you solve absolute value inequalities involving "less than"?
2. What does it mean when an absolute value inequality has no solution?
3. How do you graph solutions of absolute value inequalities on a number line?
4. What happens if you multiply or divide an inequality by a negative number?
5. Can absolute value inequalities have both "less than" and "greater than" conditions simultaneously?

**Tip:** Always be careful when dividing or multiplying inequalities by negative numbers, as it requires flipping the inequality sign.

Learn how to solve the absolute value inequality |6 - 5x| + 1 >= 9 step-by-step with detailed explanations. This problem covers absolute value inequalities and involves isolating the absolute value, considering two cases, and solving each case separately. Suitable for students in Grades 9-12.

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