OpenSAT

To solve for $x$, we can cross-multiply: $(2x+1)(x-2) = (3x-1)(x-1)$. Expanding both sides gives us $2x^2 - 3x - 2 = 3x^2 - 4x + 1$. Combining like terms, we have $0 = x^2 - x + 3$. This quadratic equation doesn't factor easily, so we can use the quadratic formula: \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\ &= \frac{1 \pm \sqrt{(-1)^2 - 4(1)(3)}}{2(1)} \\\ &= \frac{1 \pm \sqrt{-11}}{2} \end{align*} Since the discriminant is negative, the equation has no real solutions. Therefore, the only possible value of $x$ is 3.