Question

The additive inverse of $\frac{-7}{19}$ is:
(a) $\frac{-7}{19}$
(b) $\frac{7}{19}$
(c) $\frac{19}{7}$
(d) $\frac{-19}{7}$

Answer 1

We know that, if $a$ and $b$ are the additive inverse of each other, then $a+b=0$
Suppose, $x$ is the additive inverse of $\frac{-7}{19}$
Then, $x+(-\frac{7}{19})=0\Rightarrow x-\frac{7}{19}=0\Rightarrow x=\frac{7}{19}$
Hence, additive inverse of $\frac{-7}{19}$ is $\frac{7}{19}$ i.e option (b)