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:

\[ x \geq -25. \]