Answer:
Since.\[\frac{2x-1}{3}=\frac{x-2}{3}+1\] \[\left[ On\,\,transposing\frac{x-2}{3}to\,LHS \right]\] \[\begin{align} & \frac{2x-1}{3}-\frac{x-2}{3}=1 \\ & \frac{\left( 2x-1 \right)-\left( x-2 \right)}{3}=1 \\ \end{align}\] \[\begin{align} & \frac{2x-1-x+2}{3}=1 \\ & \frac{x+1}{3}=1 \\ \end{align}\] \[\begin{align} & x+1=3\times 1 \\ & x+1=3 \\ \end{align}\] Thus, \[\begin{align} & x=3-1 \\ & x=2 \\ \end{align}\]
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