Question - Solving a Compound Inequality

Given the compound inequality \( \frac{3x+1}{2} \leq 5 + 2x \) and \( 1 + 5x \geq 8x \).

First, solve \( \frac{3x+1}{2} \leq 5 + 2x \):

Multiply both sides by 2 to eliminate the fraction: \( 3x+1 \leq 10 + 4x \).

Subtract \( 3x \) from both sides: \( 1 \leq 10 + x \).

Subtract 10 from both sides: \( -9 \leq x \) or \( x \geq -9 \).

Now, solve \( 1 + 5x \geq 8x \):

Subtract \( 5x \) from both sides: \( 1 \geq 3x \).

Divide both sides by 3: \( \frac{1}{3} \geq x \) or \( x \leq \frac{1}{3} \).

Combine the solutions: \( x \geq -9 \) and \( x \leq \frac{1}{3} \).

So the solution to the compound inequality is \( -9 \leq x \leq \frac{1}{3} \).

Thus, the correct answer is D. \( -9 \leq x \leq \frac{1}{3} \).