[tex]\dfrac{2}{5}n+\dfrac{1}{10}=\dfrac{1}{2}(n+4)\ \ \ |\cdot10\\\\10\cdot\dfrac{2}{5}n+10\cdot\dfrac{1}{10}=10\cdot\dfrac{1}{2}(n+4)\\\\2\cdot2n+1=5(n+4)\ \ \ \ |\text{use distributive property}\\\\4n+1=(5)(n)+(5)(4)\\\\4n+1=5n+20\ \ \ |-1\\\\4n=5n+19\ \ \ \ |-5n\\\\-n=19\ \ \ |\cdot(-1)\\\\\boxed{n=-19}[/tex]
