(a) Start with the equation \(\log_3(2x + 1) = 1 + 2\log_3(x - 1)\).

Using the law of logarithms, \(2\log_3(x - 1) = \log_3((x - 1)^2)\).

Thus, the equation becomes \(\log_3(2x + 1) = \log_3(3) + \log_3((x - 1)^2)\).

This simplifies to \(\log_3(2x + 1) = \log_3(3(x - 1)^2)\).

Equating the arguments gives \(2x + 1 = 3(x - 1)^2\).

Expanding \(3(x - 1)^2\) gives \(3(x^2 - 2x + 1) = 3x^2 - 6x + 3\).

Thus, \(2x + 1 = 3x^2 - 6x + 3\).

Rearranging gives \(3x^2 - 8x + 2 = 0\).

(b) Use the quadratic equation from part (a) to solve for \(y\):

\(3(2y)^2 - 8(2y) + 2 = 0\) simplifies to \(12y^2 - 16y + 2 = 0\).

Divide through by 2: \(6y^2 - 8y + 1 = 0\).

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6, b = -8, c = 1\).

\(y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 6 \cdot 1}}{12}\).

\(y = \frac{8 \pm \sqrt{64 - 24}}{12}\).

\(y = \frac{8 \pm \sqrt{40}}{12}\).

\(y = \frac{8 \pm 2\sqrt{10}}{12}\).

\(y = \frac{4 \pm \sqrt{10}}{6}\).

Calculating gives \(y \approx 1.19\) (to 2 decimal places).