\(\dfrac{2x+3}{4}+\dfrac{x-1}{2}\ge\dfrac{2-x}{3}\\ \Leftrightarrow\dfrac{\left(2x+3\right).3}{12}+\dfrac{6\left(x-1\right)}{12}-\dfrac{4.\left(2-x\right)}{12}\ge0\\ \Leftrightarrow\dfrac{6x+9+6x-6-8+4x}{12}\ge0\\ \Leftrightarrow16x-5\ge0\\ \Leftrightarrow x\ge\dfrac{5}{16}\)


