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)

Solve the equation \( 3x-(6x+1) \leq 8x+2(x-3) \). Solve the inequality by following steps: - step0: Solve for \(x\): \(3x-\left(6x+1\right)\leq 8x+2\left(x-3\right)\) - step1: Subtract the terms: \(-3x-1\leq 8x+2\left(x-3\right)\) - step2: Move the expression to the left side: \(-3x-1-\left(8x+2\left(x-3\right)\right)\leq 0\) - step3: Subtract the terms: \(-11x-1-2\left(x-3\right)\leq 0\) - step4: Calculate: \(-13x+5\leq 0\) - step5: Move the constant to the right side: \(-13x\leq 0-5\) - step6: Remove 0: \(-13x\leq -5\) - step7: Change the signs: \(13x\geq 5\) - step8: Divide both sides: \(\frac{13x}{13}\geq \frac{5}{13}\) - step9: Divide the numbers: \(x\geq \frac{5}{13}\) The solution to the inequality \(3x-(6x+1) \leq 8x+2(x-3)\) is \(x \geq \frac{5}{13}\). To write this in interval notation, we can express it as \([ \frac{5}{13}, \infty)\). To graph this inequality, we can plot the point \(\frac{5}{13}\) on the number line and shade the region to the right of this point, indicating that all values of \(x\) greater than or equal to \(\frac{5}{13}\) satisfy the inequality.