Answer:
[tex]\frac{2m-1}{5}[/tex]
Step-by-step explanation:
[tex]2\left(\frac{1}{5}m-\frac{2}{5}\right)+\frac{3}{5}\\\\2\left(\frac{1}{5}m-\frac{2}{5}\right)=2\left(\frac{m}{5}-\frac{2}{5}\right)\\\\\frac{m}{5}-\frac{2}{5}=\frac{m-2}{5}\\\\=\frac{2\left(m-2\right)}{5}+\frac{3}{5}\\\\\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\\\=\frac{2\left(m-2\right)+3}{5}\\\\\mathrm{Expand}\:\left(m-2\right)\cdot \:2+3:\quad 2m-1\\\\=\frac{2m-1}{5}[/tex]
