Question

Evaluate:

(i) $\frac{3}{4}-\frac{4}{5}$
(ii) $-3-\frac{4}{7}$
(iii) $\frac{7}{24}-\frac{19}{36}$
(iv) $\frac{14}{15}-\frac{13}{20}$
(v) $\frac{4}{9}-\frac{2}{-3}$
(vi) $\frac{7}{11}-\frac{-4}{-11}$
(vii) $\frac{-5}{14}-\frac{-2}{7}$
(viii) $\frac{-5}{-8}-\frac{-3}{4}$

Answer 1

$\begin{array}{l}\text{( i)}\\ \\ \frac{3}{4}-\frac{4}{5}\\ =\frac{3}{4}-\frac{4}{5}=\frac{3}{4}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{4}{5}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\text{4 and 5 is 20.}\\ \begin{array}{l}=\frac{15-16}{20}\\ =\frac{-1}{20}\end{array}\\ \end{array}$


$\begin{array}{l}\left(ii\right)\\ \\ \frac{-3}{1}-\frac{4}{7}=\frac{-3}{1}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{4}{7}\right)\\ =\frac{-3}{1}-\frac{4}{7}\\ =\frac{-21-4}{7}\\ =\frac{-25}{7}\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\\ \frac{7}{24}-\frac{19}{36}=\frac{7}{24}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{19}{36}\right)\\ =\frac{7}{24}-\frac{19}{36}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}24\mathrm{and}36\mathrm{is}72.\\ \begin{array}{l}=\frac{21-38}{72}\\ =\frac{-17}{72}\end{array}\end{array}$

$\begin{array}{l}\left(iv\right)\\ \\ \frac{14}{15}-\frac{13}{20}=\frac{14}{15}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{13}{20}\right)\\ \\ =\frac{14}{15}-\frac{13}{20}\\ \\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}15\mathrm{and}20\mathrm{is}60.\\ \begin{array}{l}=\frac{56-39}{60}\\ \\ =\frac{17}{60}\end{array}\\ \end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\\ \\ \frac{4}{9}-\frac{2}{(-3)}\\ \text{We need a positive denominator.}\\ \frac{2}{-3}\times \frac{(-1)}{(-1)}=\frac{-2}{3}\\ \\ \frac{4}{9}-\frac{(-2)}{3}\text{=}\frac{4}{9}\text{+}\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-2}{3}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}3\mathrm{and}9\mathrm{is}9.\\ \begin{array}{l}=\frac{4+6}{9}\\ =\frac{10}{9}\end{array}\end{array}$

$\begin{array}{l}\\ \text{(vi)}\\ \\ \frac{7}{11}-\frac{(-4)}{(-11)}\\ \text{We need a positive denominator.}\\ \frac{-4}{-11}\times \frac{(-1)}{(-1)}=\frac{4}{11}\\ \\ \frac{7}{11}-\frac{4}{11}=\frac{7}{11}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{4}{11}\right)\\ =\frac{7-4}{11}\\ =\frac{3}{11}\\ \end{array}$
$$\begin{gathered} \left(\mathrm{vii}\right)\frac{-5}{14}-\frac{-2}{7}=\frac{-5}{14}+\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-2}{7}\right) \\ \frac{-5}{14}+\frac{2}{7} \\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}14\mathrm{and}7\mathrm{is}14. \\ \frac{-5+4}{14} \\ =\frac{-1}{14} \end{gathered}$$




$\begin{array}{l}\left(viii\right)\\ \\ \left(\frac{-5}{-8}\right)-\left(\frac{-3}{4}\right)\\ \\ \text{We need a positive denominator.}\\ \\ \frac{-5}{-8}\times \frac{-1}{-1}=\frac{5}{8}\\ \\ \frac{5}{8}-\frac{(-3)}{4}\text{=}\frac{5}{8}\text{+}\left(\mathrm{additive}\mathrm{inverse}\mathrm{of}\frac{-3}{4}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}8\mathrm{and}4\mathrm{is}8.\\ =\frac{5+6}{8}\\ \\ =\frac{11}{8}\end{array}$ [L.C.M. of 8 and 4 is 8.]