Solve the following linear inequalities and write the solution in interval notation.
$2(x+4)\le 3(x-1)+x$

• A. $\frac{11}{2}\ge x$
• B. $\frac{11}{2}\le x$
• C. $\frac{11}{4}\le x$
• D. $\frac{11}{4}\ge x$

Answer: (B)

Verified step by step guidance

1. Start with the given inequality: \$2\left(x+4\right) \le 3\left(x-1\right) + x$.
2. Distribute the constants inside the parentheses: \$2x + 8 \le 3x - 3 + x$.
3. Combine like terms on the right side: \$2x + 8 \le 4x - 3$.
4. Get all variable terms on one side and constants on the other by subtracting \$2x\( from both sides and adding \)3\( to both sides: \)8 + 3 \le 4x - 2x\( which simplifies to \)11 \le 2x$.
5. Divide both sides by 2 to isolate \(x\): \(\frac{11}{2} \le x\). This is the solution in inequality form, which can be written in interval notation as \(\left[\frac{11}{2}, \infty\right)\).