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Let's solve the given inequality:

\[
-8(6x - 5) \geq -16 + 32x
\]

### Step 1: Distribute the \(-8\) on the left side.

\[
-8(6x - 5) = -48x + 40
\]

So, the inequality becomes:

\[
-48x + 40 \geq -16 + 32x
\]

### Step 2: Move all the terms involving \(x\) to one side and the constants to the other side. Subtract \(32x\) from both sides:

\[
-48x - 32x + 40 \geq -16
\]

Simplifying:

\[
-80x + 40 \geq -16
\]

### Step 3: Subtract 40 from both sides:

\[
-80x \geq -16 - 40
\]

\[
-80x \geq -56
\]

### Step 4: Solve for \(x\) by dividing both sides by \(-80\). **Note:** When dividing by a negative number, you must reverse the inequality sign.

\[
x \leq \frac{-56}{-80}
\]

\[
x \leq \frac{7}{10}
\]

### Final Answer:
\[
x \leq \frac{7}{10}
\]

Would you like more details on any of the steps?

Here are 5 related questions:
1. How would the solution change if the inequality was strict (\(<\)) instead of \(\leq\)?
2. Can you solve the inequality if the \(x\)-terms are on the right side from the start?
3. How do we handle inequalities when dividing by negative numbers?
4. What happens if the inequality had multiple negative terms on both sides?
5. How would we approach the problem if there was a fraction in the original inequality?

**Tip:** Always reverse the inequality sign when multiplying or dividing by a negative number!

What values of x solve the following inequality? -8(6x - 5) >= -16 + 32x

Learn how to solve the linear inequality -8(6x - 5) >= -16 + 32x with a detailed, step-by-step solution. This problem involves key algebraic techniques, including the distributive property and handling inequalities when dividing by a negative number. Suitable for students in Grades 8-10.

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