Find: $\frac{-13}{7}+\frac{6}{7},\frac{19}{5}+\left ( \frac{-7}{5} \right )$

For the given sum:  $\frac{-13}{7}+\frac{6}{7}$

Here the denominator value is same that is 7 hence we can sum the numerator as:

$\frac{-13}{7}+\frac{6}{7}= \frac{-13+6}{7} = \frac{-7}{7} = -1$

For the given sum:  $\frac{19}{5}+\left ( \frac{-7}{5} \right )$

Here also the denominator value is the same and is equal to 5 hence we can write it as:

$\frac{19}{5}+\left ( \frac{-7}{5} \right ) = \frac{19-7}{5} = \frac{12}{5}$

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