Solve the following inequality. Write the inequality in interval notation, and graph it. \( 3 x-(6 x+1) \leq 8 x+2(x-3) \) The solution set is (Type your answer in.interval notation. Type an integer or a fraction)

To solve the inequality \( 3 x - (6 x + 1) \leq 8 x + 2(x - 3) \), we first simplify both sides. Starting with the left side: \[ 3x - (6x + 1) = 3x - 6x - 1 = -3x - 1. \] Now simplifying the right side: \[ 8x + 2(x - 3) = 8x + 2x - 6 = 10x - 6. \] Now we can rewrite the inequality: \[ -3x - 1 \leq 10x - 6. \] Next, let's isolate \(x\). Adding \(3x\) to both sides gives us: \[ -1 \leq 13x - 6. \] Adding \(6\) to both sides gets us: \[ 5 \leq 13x. \] Now divide by \(13\): \[ \frac{5}{13} \leq x. \] This can be rewritten as: \[ x \geq \frac{5}{13}. \] In interval notation, the solution set is: \[ \left[ \frac{5}{13}, \infty \right). \] Now for the graph: On a number line, you would have a solid dot at \( \frac{5}{13} \) (indicating that it’s included) and a shaded line extending to the right towards infinity. So, the complete answer is: The solution set is \( \left[ \frac{5}{13}, \infty \right) \).