Question:

(i) $\frac{3}{4}$ and $\frac{-3}{5}$

(ii) $\frac{5}{8}$ and $\frac{-7}{12}$

(iii) $\frac{-8}{9}$ and $\frac{11}{6}$

(iv) $\frac{-5}{16}$ and $\frac{7}{24}$

(v) $\frac{7}{-18}$ and $\frac{8}{27}$

(vi) $\frac{1}{-12}$ and $\frac{2}{-15}$

(vii) $-1$ and $\frac{3}{4}$

(viii) 2 and $\frac{-5}{4}$

(ix) 0 and $\frac{-2}{5}$

Solution:

1. The denominators of the given rational numbers are 4 and 5.

LCM of 4 and 5 is 20.

Now,

$\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}$ and $\frac{-3}{5}=\frac{-3 \times 4}{5 \times 4}=\frac{-12}{20}$

$\therefore \frac{3}{4}+\frac{-3}{5}=\frac{15}{20}+\frac{-12}{20}=\frac{15+(-12)}{20}=\frac{15-12}{20}=\frac{3}{20}$

2.​ The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now,

$\frac{5}{8}=\frac{5 \times 3}{8 \times 3}=\frac{15}{24}$ and $\frac{-7}{12}=\frac{-7 \times 2}{12 \times 2}=\frac{-14}{24}$

$\therefore \frac{5}{8}+\frac{-7}{12}=\frac{15}{24}+\frac{-14}{24}=\frac{15+(-14)}{24}=\frac{15-14}{24}=\frac{1}{24}$

3. ​The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now,

$\frac{-8}{9}=\frac{-8 \times 2}{9 \times 2}=\frac{-16}{18}$ and $\frac{11}{6}=\frac{11 \times 3}{6 \times 3}=\frac{33}{18}$

$\therefore \frac{-8}{9}+\frac{11}{6}=\frac{-16}{18}+\frac{33}{18}=\frac{-16+33}{18}=\frac{-16+33}{18}=\frac{17}{18}$

4.​ The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now,

$\frac{-5}{16}=\frac{-5 \times 3}{16 \times 3}=\frac{-15}{48}$ and $\frac{7}{24}=\frac{7 \times 2}{24 \times 2}=\frac{14}{48}$

$\therefore \frac{-5}{16}+\frac{7}{24}=\frac{-15}{48}+\frac{14}{48}=\frac{-15+14}{48}=\frac{-1}{48}$

5. We will first write each of the given numbers with positive denominators.

$\frac{7}{-18}=\frac{7 \times(-1)}{-18 \times(-1)}=\frac{-7}{18}$

​The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now,

$\frac{-7}{18}=\frac{-7 \times 3}{18 \times 3}=\frac{-21}{54}$ and $\frac{8}{27}=\frac{8 \times 2}{27 \times 2}=\frac{16}{54}$

$\therefore \frac{7}{-18}+\frac{8}{27}=\frac{-21}{54}+\frac{16}{54}=\frac{-21+16}{54}=\frac{-5}{54}$

6. ​We will first write each of the given numbers with positive denominators.

$\frac{1}{-12}=\frac{1 \times(-1)}{-12 \times(-1)}=\frac{-1}{12}$ and $\frac{2}{-15}=\frac{2 \times(-1)}{-15 \times(-1)}=\frac{-2}{15}$

​The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60.

Now,

$\frac{-1}{12}=\frac{-1 \times 5}{12 \times 5}=\frac{-5}{60}$ and $\frac{-2}{15}=\frac{-2 \times 4}{15 \times 4}=\frac{-8}{60}$

$\therefore \frac{1}{-12}+\frac{2}{-15}=\frac{-5}{60}+\frac{-8}{60}=\frac{-5+(-8)}{60}=\frac{-5-8}{60}=\frac{-13}{60}$

7. We can write $-1$ as $\frac{-1}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now,

$\frac{-1}{1}=\frac{-1 \times 4}{1 \times 4}=\frac{-4}{4}$ and $\frac{3}{4}=\frac{3 \times 1}{4 \times 1}=\frac{3}{4}$

$\therefore-1+\frac{3}{4}=\frac{-4}{4}+\frac{3}{4}=\frac{-4+3}{4}=\frac{-1}{4}$

8. We can write 2 as $\frac{2}{1}$.

The denominators of the given rational numbers are 1 and 4.

LCM of 1 and 4 is 4.

Now,

$\frac{2}{1}=\frac{2 \times 4}{1 \times 4}=\frac{8}{4}$ and $\frac{-5}{4}=\frac{-5 \times 1}{4 \times 1}=\frac{-5}{4}$

$\therefore 2+\frac{(-5)}{4}=\frac{8}{4}+\frac{(-5)}{4}=\frac{8+(-5)}{4}=\frac{8-5}{4}=\frac{3}{4}$

9. We can write 0 as $\frac{0}{1}$.

The denominators of the given rational numbers are 1 and 5.

LCM of 1 and 5 is 5, that is, (1 ×">× 5).

Now,

$\frac{0}{1}=\frac{0 \times 5}{1 \times 5}=\frac{0}{5}=0$ and $\frac{-2}{5}=\frac{-2 \times 1}{5 \times 1}=\frac{-2}{5}$

$\therefore 0+\frac{(-2)}{5}=\frac{0}{5}+\frac{(-2)}{5}=\frac{0+(-2)}{5}=\frac{0-2}{5}=\frac{-2}{5}$