Solve for s.
– 38+72s>–13s–93≥–96+72s

We are given the compound inequality:

\[
-38 + 72s > -13s - 93 \geq -96 + 72s
\]

This inequality consists of two parts. We'll solve each part separately.

### Part 1: Solve \(-38 + 72s > -13s - 93\)

1. Add \(13s\) to both sides to get rid of the \(s\) term on the right:
   \[
   -38 + 72s + 13s > -93
   \]
   Simplifying:
   \[
   -38 + 85s > -93
   \]

2. Add 38 to both sides:
   \[
   85s > -93 + 38
   \]
   Simplifying:
   \[
   85s > -55
   \]

3. Divide by 85:
   \[
   s > \frac{-55}{85}
   \]
   Simplifying the fraction:
   \[
   s > -\frac{11}{17}
   \]

### Part 2: Solve \(-13s - 93 \geq -96 + 72s\)

1. Add \(13s\) to both sides:
   \[
   -93 \geq -96 + 72s + 13s
   \]
   Simplifying:
   \[
   -93 \geq -96 + 85s
   \]

2. Add 96 to both sides:
   \[
   -93 + 96 \geq 85s
   \]
   Simplifying:
   \[
   3 \geq 85s
   \]

3. Divide by 85:
   \[
   \frac{3}{85} \geq s
   \]
   This can be rewritten as:
   \[
   s \leq \frac{3}{85}
   \]

### Final Answer:

We now combine both results:

\[
-\frac{11}{17} < s \leq \frac{3}{85}
\]

Thus, the solution is:

\[
s \in \left( -\frac{11}{17}, \frac{3}{85} \right]
\]

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How do we solve compound inequalities?
2. What happens when dividing both sides of an inequality by a negative number?
3. How do you graph a solution to a compound inequality on a number line?
4. What real-world problems can be modeled using compound inequalities?
5. How would this inequality change if it were absolute values instead?

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

Solve for s. 38 + 72s > –13s – 93 ≥ –96 + 72s

Discover how to solve the compound inequality 38 + 72s > –13s – 93 ≥ –96 + 72s with a detailed breakdown. This guide explores key algebraic concepts like addition, subtraction, and the transitive property of inequalities. Ideal for students in Grades 9-11.

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