let's keep in mind that we have denominators of 2, 4 and 2, and thus we can use their LCD of 4, and multiply both sides by that LCD to do away with the denominators, let's proceed,
[tex]\bf \cfrac{1}{2}n+\cfrac{3}{4}n=\cfrac{1}{2}\implies \stackrel{\textit{multipying both sides by }\stackrel{LCD}{4}}{4\left( \cfrac{1}{2}n+\cfrac{3}{4}n \right)=4\left( \cfrac{1}{2} \right)}\implies 2n+3n=2 \\\\\\ 5n=2\implies n=\cfrac{2}{5}[/tex]
