Question

Add and express the sum as a mixed fraction:
(i) $\frac{-12}{5}+\frac{43}{10}$
(ii) $\frac{24}{7}+\frac{-11}{4}$
(iii) $\frac{-31}{6}+\frac{-27}{8}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{-12}{5}+\frac{43}{10} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}5\mathrm{and}10\mathrm{is}10. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{-12}{5}\mathrm{and}\frac{43}{10}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denominator}\mathrm{as}\mathrm{LCM}. \\ \frac{-12}{5}=\frac{12\times 2}{5\times 2} \\ \frac{43}{10}=\frac{43\times 1}{10\times 1} \\ \mathrm{Therefore}, \\ \frac{-12\times 2}{5\times 2}+\frac{43}{10}=\frac{-24+43}{10}=\frac{19}{10}=1\frac{9}{10} \\ \left(\mathrm{ii}\right)\frac{24}{7}+\frac{-11}{4} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}7\mathrm{and}4\mathrm{is}28. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{24}{7}\mathrm{and}\frac{-11}{4}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denominator}\mathrm{as}\mathrm{LCM}. \\ \frac{24}{7}=\frac{24\times 4}{7\times 4} \\ \frac{-11}{4}=\frac{-11\times 7}{4\times 7} \\ \mathrm{Therefore}, \\ \frac{24\times 4}{7\times 4}+\frac{-11\times 7}{4\times 7}=\frac{96-77}{28}=\frac{19}{28} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\frac{-31}{6}+\frac{-27}{8} \\ \mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}6\mathrm{and}8\mathrm{is}24. \\ \mathrm{Now},\mathrm{we}\mathrm{express}\frac{-31}{6}\mathrm{and}\frac{-27}{8}\mathrm{into}\mathrm{forms}\mathrm{in}\mathrm{which}\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{denominator}\mathrm{as}\mathrm{LCM}. \\ \frac{-31}{6}=\frac{-31\times 4}{6\times 4} \\ \frac{-27}{8}=\frac{-27\times 3}{8\times 3} \\ \mathrm{Therefore}, \\ \frac{-31\times 4}{6\times 4}+\frac{-27\times 3}{8\times 3}=\frac{-124-81}{24}=\frac{-205}{24} \end{gathered}$$