# Solve the following

Question:

Subtract:

(i) $\frac{3}{4}$ from $\frac{1}{3}$

(ii) $\frac{-5}{6}$ from $\frac{1}{3}$

(iii) $\frac{-8}{9}$ from $\frac{-3}{5}$

(iv) $\frac{-9}{7}$ from $-1$

(v) $\frac{-18}{11}$ from 1

(vi) $\frac{-13}{9}$ from 0

(vii) $\frac{-32}{13}$ from $\frac{-6}{5}$

(viii) $-7$ from $\frac{-4}{7}$

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{13}{9}$

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

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

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

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

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

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