For [tex]A+\frac{2}{a}-\frac{5}{a}+\frac{1}{a^2}-8a+15[/tex]
Combine the fractions [tex] \frac{2}{a} - \frac{5}{a} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{2-5}{a} [/tex]
[tex]\mathrm{Subtract\:the\:numbers:}\:2-5=-3 \ \textgreater \ \frac{-3}{a} [/tex]
[tex]\mathrm{Apply\:the\:fraction\:rule}:\ \frac{-a}{b}=-\frac{a}{b} \ \textgreater \ -\frac{3}{a}[/tex]
Therefore our final answer is [tex]A-\frac{3}{a}+\frac{1}{a^2}-8a+15[/tex]
Hope this helps!
