$\text{m}-\frac{\text{m}-1}{2}=1-\frac{\text{m}-2}{3}$
$=\frac{2\text{m}-\text{m}+1}{2}=\frac{3-\text{m}+2}{3}$
$=\frac{\text{m}+1}{2}=\frac{5-\text{m}}{3}$
$=\frac{\text{m}+1}{2}=\frac{5}{3}-\frac{\text{m}}{3}$
$=\frac{\text{m}}{2}+\frac12=\frac53-\frac{\text{m}}{3}$
Transposing $\frac{\text{m}}{3}$ to L.H.S. and $\frac12$ to R.H.S., we get
$=\frac{\text{m}}{2}+-\frac{\text{m}}{3}=\frac{5}{3}-\frac12$
$=\frac{3\text{m}+2\text{m}}{6}=\frac{10-3}{6}$
Multiplying both sides by 6, we get
$=\frac{5\text{m}}{6}\times6=\frac{7}{6}\times6$
$=5\text{m}=7$
Dividing both sides by 5, we get
$=\frac{5\text{m}}{5}=\frac75$
$=\text{m}=\frac75$
Verification:
Substituting $\text{m}=\frac75$ on both sides, we get
$\frac75-\frac{7-5}{10}=1-\frac{7-10}{15}$
$\frac75-\frac{2}{10}=\frac{15+3}{15}$
$\frac{14-2}{10}=\frac{15+3}{15}$
$\frac{12}{10}=\frac{18}{15}$
$\frac{6}{5}=\frac{6}{5}$
$\text{L.H.S.} =\text{R.H.S.}$
Hence, verified.
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