prove that

Given $m-\frac{m-1}{2}=1-\frac{m-2}{3}$

$\Rightarrow$ $\frac{2 m-(m-1)}{2}=\frac{3-(m-2)}{3}$

$\Rightarrow$ $3(2 m-m+1)=2(3-m+2)$ [by cross multiplication]

$\Rightarrow \quad 3(m+1)=2(5-m)$

$\Rightarrow \quad 3 m+3=10-2 m$

$\Rightarrow$ $3 m+2 m=10-3$ [transposing $-2 m$ to LHS and 3 to RHS]

$\Rightarrow$ $5 m=7$

$\Rightarrow$ $\frac{5 m}{5}=\frac{7}{5}$ [dividing both sides by 5 ]

$\therefore$ $m=\frac{7}{5}$