Question

Subtract:
(i) $\frac{3}{4}\mathrm{from}\frac{1}{3}$
(ii) $\frac{-5}{6}\mathrm{from}\frac{1}{3}$
(iii) $\frac{-8}{9}\mathrm{from}\frac{-3}{5}$
(iv) $\frac{-9}{7}\mathrm{from}-1$
(v) $\frac{-18}{11}\mathrm{from}1$
(vi) $\frac{-13}{9}\mathrm{from}0$
(vii) $\frac{-32}{13}\mathrm{from}\frac{-6}{5}$
(viii) $-7\mathrm{from}\frac{-4}{7}$

Answer 1

(i) $\left(\frac{1}{3}-\frac{3}{4}\right)=\frac{1}{3}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{3}{4}\right)$

= $\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{-3}{4}}\right)=\left({\displaystyle \frac{4}{12}}+{\displaystyle \frac{-9}{12}}\right)=\left(\frac{4-9}{12}\right)=\frac{-5}{12}$


(ii) $\left(\frac{1}{3}-\frac{-5}{6}\right)=\frac{1}{3}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-5}{6}\right)$

= $\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{5}{6}}\right)$ (Because the additive inverse of $\frac{-5}{6}\mathrm{is}\frac{5}{6}$)

=$\left({\displaystyle \frac{2}{6}}+{\displaystyle \frac{5}{6}}\right)=\left(\frac{2+5}{6}\right)=\frac{7}{6}$


(iii) $\left(\frac{-3}{5}-\frac{-8}{9}\right)=\frac{-3}{5}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-8}{9}\right)$

= $\left({\displaystyle \frac{-3}{5}}+{\displaystyle \frac{8}{9}}\right)$ (Because the additive inverse of $\frac{-8}{9}\mathrm{is}\frac{8}{9}$)

=$\left({\displaystyle \frac{-27}{45}}+{\displaystyle \frac{40}{45}}\right)=\left(\frac{-27+40}{45}\right)=\frac{13}{45}$


(iv) $\left(-1-\frac{-9}{7}\right)=-1+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-9}{7}\right)$

=$\left({\displaystyle \frac{-1}{1}}+{\displaystyle \frac{9}{7}}\right)$ (Because the additive inverse of $\frac{-9}{7}\mathrm{is}\frac{9}{7}$)

=$\left({\displaystyle \frac{-7}{7}}+{\displaystyle \frac{9}{7}}\right)=\left(\frac{-7+9}{7}\right)=\frac{2}{7}$


(v) $\left(1-\frac{-18}{11}\right)=1+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-18}{11}\right)$

=$\left({\displaystyle \frac{1}{1}}+{\displaystyle \frac{18}{11}}\right)$ (Because the additive inverse of $\frac{-18}{11}\mathrm{is}\frac{18}{11}$)

= $\left({\displaystyle \frac{11}{11}}+{\displaystyle \frac{18}{11}}\right)=\left(\frac{11+18}{11}\right)=\frac{29}{11}$


(vi) $\left(0-\frac{-13}{9}\right)=0+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-13}{9}\right)$

=$\left(0+{\displaystyle \frac{13}{9}}\right)$ (Because the additive inverse of $\frac{-13}{9}\mathrm{is}\frac{13}{9}$)

=$\frac{13}{9}$

(vii) $\left(\frac{-6}{5}-\frac{-32}{13}\right)=\frac{-6}{5}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-32}{13}\right)$

=$\left({\displaystyle \frac{-6}{5}}+{\displaystyle \frac{32}{13}}\right)$ (Because the additive inverse of $\frac{-32}{13}\mathrm{is}\frac{32}{13}$)

= $\left({\displaystyle \frac{-78}{65}}+{\displaystyle \frac{160}{65}}\right)=\left(\frac{-78+160}{65}\right)=\frac{82}{65}$


(viii) $\left(\frac{-4}{7}-\frac{-7}{1}\right)=\frac{-4}{7}+\left(\mathrm{Additive}\mathrm{inverse}\mathrm{of}\frac{-7}{1}\right)$

= $\left({\displaystyle \frac{-4}{7}}+{\displaystyle \frac{7}{1}}\right)$ (Because the additive inverse of $\frac{-7}{1}\mathrm{is}\frac{7}{1}$)

= $\left({\displaystyle \frac{-4}{7}}+{\displaystyle \frac{49}{7}}\right)=\left(\frac{-4+49}{7}\right)=\frac{45}{7}$