## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\dfrac{7}{12}$
RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $a^1=a$ Use the rule above to obtain: $3^{-1}+4^{-1} \\=\dfrac{1}{3^1}+ \dfrac{1}{4^1} \\=\dfrac{1}{3} + \dfrac{1}{4}$ Make the fractions similar using their LCD of $12$ to obtain: $=\dfrac{1\color{blue}{(4)}}{3\color{blue}{(4)}} +\dfrac{1\color{blue}{(3)}}{4\color{blue}{(3)}} \\=\dfrac{4}{12}+\dfrac{3}{12}$ Add the numerators and copy/retain the denominator to obtain: $=\dfrac{4+3}{12} \\=\dfrac{7}{12}$