To solve the compound inequality \(6c + 8 \leq -10\) or \(8c + 12 > 20\), we will solve each inequality separately.
First Inequality: \(6c + 8 \leq -10\)
- Start by isolating \(c\). We will first subtract 8 from both sides: \[ 6c + 8 - 8 \leq -10 - 8 \] Simplifying gives: \[ 6c \leq -18 \]
- Next, divide both sides by 6: \[ c \leq -3 \]
Second Inequality: \(8c + 12 > 20\)
- Isolate \(c\) by first subtracting 12 from both sides: \[ 8c + 12 - 12 > 20 - 12 \] Simplifying gives: \[ 8c > 8 \]
- Now, divide both sides by 8: \[ c > 1 \]
Combining Solutions:
From the solutions obtained, we have:
- From the first inequality: \(c \leq -3\)
- From the second inequality: \(c > 1\)
Since this is a compound inequality with "or," the solution includes both parts.
Thus, the solutions to the compound inequality \(6c + 8 \leq -10\) or \(8c + 12 > 20\) are: \[ c \leq -3 \quad \text{or} \quad c > 1 \]
Final Answer: The solution set is \(c \leq -3\) or \(c > 1\).