Answer
$\dfrac{7}{12}$
Work Step by Step
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:
$=\dfrac{4+3}{12}
\\=\dfrac{7}{12}$
