(3y-1)(2y+1)>(2y-1)(2+3y)

To solve the inequality \((3y - 1)(2y + 1) > (2y - 1)(2 + 3y)\), we will first expand both sides.

Step 1: Expand both sides

1. Left Side: \[ (3y - 1)(2y + 1) = 3y(2y) + 3y(1) - 1(2y) - 1(1) = 6y^2 + 3y - 2y - 1 = 6y^2 + y - 1 \]

2. Right Side: \[ (2y - 1)(2 + 3y) = 2y(2) + 2y(3y) - 1(2) - 1(3y) = 4y + 6y^2 - 2 - 3y = 6y^2 + y - 2 \]

Step 2: Set up the inequality

Now we can rewrite the inequality: \[ 6y^2 + y - 1 > 6y^2 + y - 2 \]

Step 3: Simplify the inequality

Subtract \(6y^2 + y\) from both sides: \[ -1 > -2 \]

This statement is always true.

Step 4: Conclusion

Since \(-1 > -2\) holds for all \(y\), the given inequality \((3y - 1)(2y + 1) > (2y - 1)(2 + 3y)\) is true for all real numbers \(y\).

Thus, the solution set is: \[ y \in \mathbb{R} \quad \text{(all real numbers)} \]