To solve the inequality \( \frac{2x + 3}{3} < \frac{3x - 1}{2} \), we can start by eliminating the fractions. To do this, we find a common denominator, which in this case is 6.
Multiplying both sides by 6 gives:
\[ 6 \cdot \frac{2x + 3}{3} < 6 \cdot \frac{3x - 1}{2} \]
This simplifies to:
\[ 2(2x + 3) < 3(3x - 1) \]
Expanding both sides results in:
\[ 4x + 6 < 9x - 3 \]
Next, we will move the terms involving \( x \) to one side and the constant terms to the other side. Subtract \( 4x \) from both sides:
\[ 6 < 9x - 4x - 3 \]
This simplifies to:
\[ 6 < 5x - 3 \]
Next, add 3 to both sides:
\[ 6 + 3 < 5x \]
This results in:
\[ 9 < 5x \]
Now, divide both sides by 5:
\[ \frac{9}{5} < x \]
This is equivalent to:
\[ x > \frac{9}{5} \]
In interval notation, the solution is:
\[ \left( \frac{9}{5}, \infty \right) \]
Thus, the solution to the inequality \( \frac{2x + 3}{3} < \frac{3x - 1}{2} \) is \( x > \frac{9}{5} \).