Question

(i) $\frac{3}{4}\mathrm{and}\frac{-3}{5}$
(ii) $-3\mathrm{and}\frac{3}{5}$
(iii) $\frac{-7}{27}\mathrm{and}\frac{11}{18}$
(iv) $\frac{31}{-4}\mathrm{and}\frac{-5}{8}$

Answer 1

(i)

$$\begin{gathered} \frac{3}{4}+\frac{-3}{5} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}4\mathrm{and}5\mathrm{is}20. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{3}{4}\mathrm{and}\frac{-3}{5}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denominator}20. \\ \frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20} \\ \frac{-3}{5}=\frac{-3\times 4}{5\times 4}=\frac{-12}{20} \\ \mathrm{Therefore},\frac{3}{4}+\frac{-3}{5}=\frac{15}{20}+\frac{-12}{20}=\frac{15-12}{20}=\frac{3}{20} \end{gathered}$$

(ii)

$$\begin{gathered} -3+\frac{3}{5} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}1\mathrm{and}5\mathrm{is}5. \\ \mathrm{Now},\mathrm{we}\mathrm{express}-3\mathrm{and}\frac{3}{5}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denominator}5. \\ \frac{-3}{1}=\frac{3\times 5}{1\times 5}=\frac{-15}{5} \\ \frac{3}{5}=\frac{3\times 1}{5\times 1}=\frac{3}{5} \\ \mathrm{Therefore},-3+\frac{3}{5}=\frac{-15}{5}+\frac{3}{5}=\frac{-12}{5} \end{gathered}$$

(iii)

$$\begin{gathered} \frac{-7}{27}+\frac{11}{18} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}27\mathrm{and}18\mathrm{is}54. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{-7}{27}\mathrm{and}\frac{11}{18}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denomiator}54. \\ \frac{-7}{27}=\frac{-7\times 2}{27\times 2}=\frac{-14}{54} \\ \frac{11}{18}=\frac{11\times 3}{18\times 3}=\frac{33}{54} \\ \mathrm{Therefore},\frac{-7}{27}+\frac{11}{18}=\frac{-14}{54}+\frac{33}{54}=\frac{-14+33}{54}=\frac{19}{54} \end{gathered}$$

(iv)

$$\begin{gathered} \frac{-31}{4}+\frac{-5}{8} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}4\mathrm{and}8\mathrm{is}8. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{-31}{4}\mathrm{and}\frac{-5}{8}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denomiator}8. \\ \frac{-31}{4}=\frac{-31\times 2}{4\times 2}=\frac{-62}{8} \\ \frac{-5}{8}=\frac{-5\times 1}{8\times 1}=\frac{-5}{8} \\ \mathrm{Therefore},\frac{-31}{4}+\frac{-5}{8}=\frac{-62}{8}+\frac{-5}{8}=\frac{-62-5}{8}=\frac{-67}{8} \end{gathered}$$