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) $-7\mathrm{from}\frac{-4}{7}$
(ix) $\frac{5}{9}\mathrm{from}\frac{-2}{3}$
(x) $5\mathrm{from}\frac{-3}{5}$

Answer 1

$\begin{array}{l}\left(i\right)\\ \frac{1}{3}-\frac{3}{4}=\frac{1}{3}+(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{3}{4})\\ \text{L.C.M.}\text{of}\text{3 and 4 is 12.}\\ \frac{1}{3}+\left(-\frac{3}{4}\right)=\frac{4+(-9)}{12}=\frac{-5}{12}\end{array}$


$\begin{array}{l}\left(ii\right)\\ \\ \frac{1}{3}-\frac{(-5)}{6}=\frac{1}{3}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-5}{6}\right)\\ =\frac{1}{3}+\frac{5}{6}\\ \mathrm{L.C.M}.\text{of 3 and 6 is 6.}\\ \begin{array}{l}=\frac{2+5}{6}\\ =\frac{7}{6}\end{array}\end{array}$

$\begin{array}{l}\left(iii\right)\\ \\ \frac{(-3)}{5}-\left(\frac{-8}{9}\right)=\frac{(-3)}{5}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-8}{9}\right)\\ =\frac{(-3)}{5}+\frac{8}{9}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}5\mathrm{and}9\mathrm{is}45.\\ \begin{array}{l}=\frac{-27+40}{45}\\ =\frac{13}{45}\end{array}\end{array}$

$\begin{array}{l}\left(iv\right)\\ \\ \frac{(-1)}{1}-\frac{(-9)}{7}=\frac{(-1)}{1}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-9}{7}\right)\\ =\frac{(-1)}{1}+\frac{9}{7}\\ L.C.M.of\; 7\; and\; 1\; is\; 7.\\ \begin{array}{l}=\frac{-7+9}{7}\\ =\frac{2}{7}\end{array}\end{array}$

$\begin{array}{l}\left(v\right)\\ \\ \frac{1}{1}-\frac{(-18)}{11}=\frac{1}{1}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-18}{11}\right)\\ =\frac{\left(1\right)}{1}+\frac{18}{11}\\ =\frac{11+18}{11}\\ =\frac{29}{11}\end{array}$

(vi)
$$\begin{gathered} 0-\left(\frac{-13}{9}\right)=0+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-13}{9}\right) \\ =0+\frac{13}{9} \\ =\frac{13}{9} \end{gathered}$$

.$\begin{array}{l}\left(\mathrm{vii}\right)\\ \\ \frac{(-6)}{5}-\frac{(-32)}{13}=\frac{(-6)}{5}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-32}{13}\right)\\ =\frac{-6}{5}+\frac{32}{13}\\ L.C.M.of5\mathrm{and}13\mathrm{is}65\\ \begin{array}{l}=\frac{-78+160}{65}\\ =\frac{82}{65}\end{array}\end{array}$


$$\begin{gathered} \begin{array}{l}\left(\mathrm{viii}\right)\\ \\ \\ \begin{array}{l}\begin{array}{l}\frac{-4}{7}-\frac{(-7)}{1}=\frac{-4}{7}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-7}{1}\right)\\ =\frac{(-4)}{7}+\frac{7}{1}\\ \text{L.C.M. of 7 and 1 is 7.}\end{array}\\ =\frac{-4+49}{7}\end{array}\end{array} \\ =\frac{45}{7} \end{gathered}$$
$\begin{array}{l}\left(\mathrm{ix}\right)\\ \\ \\ \frac{-2}{3}-\frac{5}{9}=\frac{-2}{3}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{5}{9}\right)\\ =\frac{-2}{3}-\frac{5}{9}\\ \text{L.C.M. of}\mathrm{3\; and\; 9\; is\; 9}.\\ \begin{array}{l}=\frac{-6-5}{9}\\ =\frac{-11}{9}\end{array}\end{array}$

$\begin{array}{l}\text{(x)}\\ \\ \frac{-3}{5}-\frac{5}{1}=\frac{-3}{5}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{5}{1}\right)\\ =\frac{-3}{5}-\frac{5}{1}\\ \text{L.C.M. of}\mathrm{5\; and\; 1\; is\; 5}.\\ \begin{array}{l}=\frac{-3-25}{5}\\ =\frac{-28}{5}\end{array}\end{array}$