Solve the inequality. Express the solution set in interval notation and graph. 
$\(\frac\)13\(\left\)(x+1\(\right\))\(\ge\]\frac\)15\(\left\)(3+2x\(\right\))$

• A. $\(\left\)(-4,\(\infty\]\right\))$
• B. $\(\left\)[4,\(\infty\]\right\))$
• C. $\(\left\)[-4,\(\infty\]\right\))$
• D. $\(\left\)(-\(\infty\),-4]\(\right\).$

Answer: (D)

Verified step by step guidance

1. Start by distributing the constants in the inequality: \(13(x+1) \geq 15(3+2x)\). This becomes \(13x + 13 \geq 45 + 30x\).
2. Next, move all terms involving \(x\) to one side of the inequality. Subtract \(13x\) from both sides to get \(13 \geq 45 + 17x\).
3. Subtract 45 from both sides to isolate the term with \(x\): \(13 - 45 \geq 17x\), which simplifies to \(-32 \geq 17x\).
4. Divide both sides by 17 to solve for \(x\): \(-\frac{32}{17} \geq x\). This can be rewritten as \(x \leq -\frac{32}{17}\).
5. Express the solution in interval notation. Since \(x\) is less than or equal to \(-\frac{32}{17}\), the interval is \((-\infty, -\frac{32}{17}]\). The graph of this solution is a line extending to the left from \(-\frac{32}{17}\) with a closed circle at \(-\frac{32}{17}\), indicating that \(-\frac{32}{17}\) is included in the solution set.