It is given that,

The sum of the two numbers = -8

One of the number = \(\frac{-15}{7}\)

Suppose the other number is x

Since,

The sum is -8

Therefore,

x + \(\frac{-15}{7}\) = -8

\(\frac{7x-15}{7}\) = -8

7x – 15 = -56

7x = -56 + 15

7x = -41

x = \(\frac{-41}{7}\)

Let the second number be ‘x’ say

\(\Rightarrow x\,+\,(\frac{-5}6)=8\)

\(\Rightarrow 8\,+\,\frac{5}6=\frac{48\,+\,5}{6}=\frac{53}{6}\)

∴ The other number (x) = \(\frac{53}{6}\)

Given sum of two numbers is -8

One of them is (-15/7)

Let the required number be x

x + (-15/7) = -8

The LCM of 7 and 1 is 7

Consider (-8/1) = (-8/1) × (7/7) = (-56/7)

On substituting

x + (-15/7) = (-56/7)

x = (-56/7) – (-15/7)

x = (-56/7) + (15/7)

x = (-56 + 15)/7

x = (-41/7)