\(\dfrac{-3}{5}+\dfrac{-1}{3}=-\dfrac{14}{15}\)
\(\dfrac{-2}{13}+\dfrac{-11}{26}=\dfrac{-15}{26}\)
\(\dfrac{-2}{3}+\dfrac{-1}{5}=\dfrac{-13}{15}\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)