i) Using the commutativity law, the addition of rational numbers is commutative.

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

let us consider the given fraction

\(\frac{-11}{5}\) and \(\frac{4}{7}\) as

\(\frac{-11}{5}\)+ \(\frac{4}{7}\)and \(\frac{4}{7}\)+\(\frac{-11}{5}\)

The denominators are 5 and 7.The LCM of 5 and 7 is 35
We rewrite the given fraction in order to get the same denominator.We get

Now,\(\frac{-11}{5}\) = \(\frac{(11\times7)}{(5\times7)}\)

 \(=\frac{(-11\times7)}{(5\times7)}\\=\frac{-77}{35}\\\frac{4}{7}=\frac{(4\times5)}{(7\times5)}\\=\frac{20}{35}\)

Since the denominators are same we can add them directly

\(\frac{-77}{35}+\frac{20}{35}\\=\frac{(-77+20)}{35}\\=\frac{-50}{35}\)

ii) Firstly we need to convert the denominators to positive numbers.We get

\(\frac{7}{-12}=\frac{(7\times-1)}{(-12\times-1)}\\=\frac{-7}{12}\)

Using the commutativity law, the addition of rational numbers is commutative.

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

In order to verify the above property let us consider the given fraction.

\(\frac{4}{9}\) and \(\frac{-7}{12}\) as

\(\frac{4}{9}\)+\(\frac{-7}{12}\) and \(\frac{-7}{12}\)+ \(\frac{4}{9}\)

The denominators are 9 and 12. By taking LCM for 9 and 12 is 36
We rewrite the given fraction in order to get the same denominator.We get

Now , \(\frac{4}{9}=\frac{(4\times4)}{(9\times4)}\\=\frac{16}{36}\)

\(\frac{-7}{12}\) = \(\frac{(-7\times3)}{(12\times3)}=\frac{-21}{36}\)

= \(\frac{-7}{12}\)+\(\frac{4}{9}\)

The denominators are 12 and 9 By taking LCM for 12 and 9 is 36.
We rewrite the given fraction in order to get the same denominator.We get

\(\frac{-7}{12}=\frac{(-7\times3)}{(12\times3) }\\=\frac{-21}{36}\)

\(\frac{4}{9}=\frac{(4\times4)}{9\times4)}\\=\frac{16}{36}\)

Since the denominators are same we can add them directly

\(\frac{-21}{36}+\frac{16}{36}=\frac{(-21+16)}{36}\\=\frac{-5}{36}\)

\(\therefore\frac{4}{9}+\frac{-7}{12}=\frac{-7}{12}+\frac{4}{9}\)  is satisfied

iii) Firstly we need to convert the denominators to positive numbers.

\(\frac{4}{-7}=\frac{(4\times-1)}{(-7\times-1)}\)

= \(\frac{-4}{7}\)

By using the commutativity law, the addition of rational numbers is commutative

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

In order to verify the above property let us consider the given fraction

\(\frac{-4}{1}\) and \(\frac{-4}{7}\) as

\(\frac{-4}{1}\)+ \(\frac{-4}{7}\) and \(\frac{-4}{7}\) + \(\frac{-4}{1}\)

The denominators are 1 and 7 By taking LCM for 1 and 7 is 7.
We rewrite the given fraction in order to get the same denominator.We get

Now,

\(\frac{-4}{1}=\frac{(-4\times7)}{(1\times7)}\\=\frac{-28}{7}\)

\(\frac{-4}{7}=\frac{(-4\times1)}{(7\times1)}\\=\frac{-4}{7}\)

Since the denominators are same we can add them directly
\(\frac{-28}{7}+\frac{-4}{7}=\frac{(-28+(-4))}{7}\\=\frac{(-28-7)}{7}\\=\frac{-32}{7}\)

\(\frac{-4}{7}+-\frac{4}{1}\)

The denominators are 7 and 1. By taking LCM for 7 and 1 is 7. We rewrite the given fraction in order to get the same denominator.

Now,

\(\frac{-4}{7}=\frac{(-4\times1)}{(7\times1)}\\=\frac{-4}{7}\)

\(\frac{-4}{1}=\frac{(-4\times7)}{(1\times7)}\\=\frac{-28}{7}\)

Since the denominators are same we can add them directly.We get

\(\frac{-4}{7}+\frac{-28}{7}=\frac{(-4+(-28))}{7}\\=\frac{(-4-28)}{7}\\=\frac{-32}{7}\)

\(\therefore \frac{-4}{1}+\frac{-4}{7}=\frac{-4}{7}+\frac{-4}{1}\) is satisfied