Solve by graphing: \( \frac{3}{2 x-1}=\frac{x+2}{x} \) \( \left\{\begin{array}{l}\{-1,0,1\} \\ \{1\} \\ \{-2,3\} \\ \{-1,1\}\end{array}\right. \)

Solve the equation \( \frac{3}{2x-1}=\frac{x+2}{x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3}{2x-1}=\frac{x+2}{x}\) - step1: Find the domain: \(\frac{3}{2x-1}=\frac{x+2}{x},x \in \left(-\infty,0\right)\cup \left(0,\frac{1}{2}\right)\cup \left(\frac{1}{2},+\infty\right)\) - step2: Cross multiply: \(3x=\left(2x-1\right)\left(x+2\right)\) - step3: Expand the expression: \(3x=2x^{2}+3x-2\) - step4: Cancel equal terms: \(0=2x^{2}-2\) - step5: Swap the sides: \(2x^{2}-2=0\) - step6: Move the constant to the right side: \(2x^{2}=2\) - step7: Divide both sides: \(\frac{2x^{2}}{2}=\frac{2}{2}\) - step8: Divide the numbers: \(x^{2}=1\) - step9: Simplify the expression: \(x=\pm \sqrt{1}\) - step10: Simplify: \(x=\pm 1\) - step11: Separate into possible cases: \(\begin{align}&x=1\\&x=-1\end{align}\) - step12: Check if the solution is in the defined range: \(\begin{align}&x=1\\&x=-1\end{align},x \in \left(-\infty,0\right)\cup \left(0,\frac{1}{2}\right)\cup \left(\frac{1}{2},+\infty\right)\) - step13: Find the intersection: \(\begin{align}&x=1\\&x=-1\end{align}\) - step14: Rewrite: \(x_{1}=-1,x_{2}=1\) The solutions to the equation \( \frac{3}{2x-1}=\frac{x+2}{x} \) are \( x = -1 \) and \( x = 1 \). Therefore, the solutions are \( \{-1,1\} \).