2 x + 1 7 ≥ x − 3 4  7 2x+1 ​ ≥ 4 x−3 ​ .

To solve the inequality \[ \frac{2x + 1}{7} \geq \frac{x - 3}{4}, \] we will first eliminate the fractions by finding a common denominator. The least common multiple of \(7\) and \(4\) is \(28\). Now, multiply both sides by \(28\): \[ 28 \cdot \frac{2x + 1}{7} \geq 28 \cdot \frac{x - 3}{4}. \] This simplifies to: \[ 4(2x + 1) \geq 7(x - 3). \] Now distribute: \[ 8x + 4 \geq 7x - 21. \] Next, subtract \(7x\) from both sides: \[ 8x - 7x + 4 \geq -21. \] This simplifies to: \[ x + 4 \geq -21. \] Now, subtract \(4\) from both sides: \[ x \geq -21 - 4, \] which simplifies to: \[ x \geq -25. \] Thus, the solution to the inequality is: \[ x \geq -25. \]