Verify the following:

Solution:

1. $\mathrm{LHS}=\frac{-12}{5}+\frac{2}{7}$

LCM of 5 and 7 is 35 .

$\frac{-12 \times 7}{5 \times 7}+\frac{2 \times 5}{7 \times 5}=\frac{-84}{35}+\frac{10}{35}=\frac{-84+10}{35}=\frac{-74}{35}$

$\mathrm{RHS}=\frac{2}{7}+\frac{-12}{5}$

LCM of 5 and 7 is 35.

$\frac{2 \times 5}{7 \times 5}+\frac{-12 \times 7}{5 \times 7}=\frac{10}{35}+\frac{-84}{35}=\frac{10-84}{35}=\frac{-74}{35}$

$\therefore \frac{-12}{5}+\frac{2}{7}=\frac{2}{7}+\frac{-12}{5}$

2. $\mathrm{LHS}=\frac{-5}{8}+\frac{-9}{13}$

LCM of 8 and 13 is 104 .

$\frac{-5 \times 13}{8 \times 13}+\frac{-9 \times 8}{13 \times 8}=\frac{-65}{104}+\frac{-72}{104}=\frac{-65+(-72)}{104}=\frac{-65-72}{104}=\frac{-137}{104}$

$\mathrm{RHS}=\frac{-9}{13}+\frac{-5}{8}$

LCM of 13 and 8 is 104.

$\frac{-9 \times 8}{13 \times 8}+\frac{-5 \times 13}{8 \times 13}=\frac{-72}{104}+\frac{-65}{104}=\frac{-72-65}{104}=\frac{-137}{104}$

$\therefore \frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$

3. $L H S=\frac{3}{1}+\frac{-7}{12}$

LCM of 1 and 12 is 12.

$\frac{3 \times 12}{1 \times 12}+\frac{-7 \times 1}{12 \times 1}=\frac{36}{12}+\frac{-7}{12}=\frac{36+(-7)}{12}=\frac{36-7}{12}=\frac{29}{12}$'

$\mathrm{RHS}=\frac{-7}{12}+\frac{3}{1}$

LCM of 12 and 1 is 12.

$\frac{-7 \times 1}{12 \times 1}+\frac{3 \times 12}{1 \times 12}=\frac{-7}{12}+\frac{36}{12}=\frac{-7+36}{12}=\frac{29}{12}$

$\therefore 3+\frac{-7}{12}=\frac{-7}{12}+3$

4. LHS $=\frac{2}{-7}+\frac{12}{-35}$

We will write the given numbers with positive denominators.

$\frac{2}{-7}=\frac{2 \times(-1)}{-7 \times(-1)}=\frac{-2}{7}$ and $\frac{12}{-35}=\frac{12 \times(-1)}{-35 \times(-1)}=\frac{-12}{35}$

LCM of 7 and 35 is 35.

$\frac{-2 \times 5}{7 \times 5}+\frac{-12 \times 1}{35 \times 1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+(-12)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$

$\mathrm{RHS}=\frac{12}{-35}+\frac{2}{-7}$

We will write the given numbers with positive denominators.

$\frac{12}{-35}=\frac{12 \times(-1)}{-35 \times(-1)}=\frac{-12}{35}$ and $\frac{2}{-7}=\frac{2 \times(-1)}{-7 \times(-1)}=\frac{-2}{7}$

LCM of 35 and 7 is 35.

$\frac{-2 \times 5}{7 \times 5}+\frac{-12 \times 1}{35 \times 1}=\frac{-10}{35}+\frac{-12}{35}=\frac{-10+(-12)}{35}=\frac{-10-12}{35}=\frac{-22}{35}$

$\therefore \frac{2}{-7}+\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}$