Question

Solve the inequality.
1. 3a + 6 less than or equal to -5(s+2)
2. -5r + 6 less than or equal to -5(r+2)

Answer 1

To solve the inequalities, we'll isolate the variable on one side of the inequality.

### 1. \( 3a + 6 \leq -5(s + 2) \)

First, distribute \(-5\) on the right side:

\[
3a + 6 \leq -5s - 10
\]

Now, we can move \(6\) to the right side:

\[
3a \leq -5s - 10 - 6
\]

This simplifies to:

\[
3a \leq -5s - 16
\]

Now, divide both sides by \(3\):

\[
a \leq \frac{-5s - 16}{3}
\]

### 2. \( -5r + 6 \leq -5(r + 2) \)

First, expand the right side:

\[
-5r + 6 \leq -5r - 10
\]

Now, add \(5r\) to both sides:

\[
6 \leq -10
\]

Since \(6\) is not less than or equal to \(-10\), this inequality is false. The solution is that there are no values of \(r\) that satisfy the inequality.

### Summary of Solutions
1. \( a \leq \frac{-5s - 16}{3} \)
2. No solutions for \( r \) (the inequality is always false).