Question

Simplify:

(i) $\frac{-8}{15}+\frac{2}{-3}$
(ii) $\frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}$
(iii) $-1+\frac{7}{-9}+\frac{11}{12}$
(iv) $\frac{-11}{39}+\frac{5}{26}+2$
(v) $2+\frac{-1}{2}+\frac{-3}{4}$
(vi) $\frac{-9}{11}+\frac{2}{3}+\frac{-3}{4}$

Answer 1

$\begin{array}{l}\left(i\right)\\ \frac{-8}{15}+\frac{2}{-3}\\ \\ \text{We need a positive denominator.}\\ \therefore \frac{2}{-3}\times \frac{-1}{-1}=\frac{-2}{3}\\ \text{Now, L.C.M. of 15 and 3 is 15.}\\ \frac{-8}{15}=\frac{-8\times 1}{15\times 1}=\frac{-8}{15}\\ \\ \frac{-2}{3}=\frac{-2\times 5}{3\times 5}=\frac{-10}{15}\\ \\ \mathrm{Now},\frac{-8}{15}+\frac{-10}{15}\\ =\frac{-8-10}{15}\\ \\ =\frac{-18}{15}\\ =\frac{-6}{5}\end{array}$

$\begin{array}{l}\left(ii\right)\\ \frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}\\ \\ \text{We need a}\text{positive denominator.}\\ \frac{13}{-15}\times \frac{-1}{-1}=\frac{-13}{15}\\ \text{Now},\text{L.C.M. of 1}0,\text{15 and 2}0\text{is 6}0.\\ \therefore \frac{-7}{10}=\frac{-7\times 6}{10\times 6}=\frac{-42}{60}\\ \\ \frac{-13}{15}=\frac{-13\times 4}{15\times 4}=\frac{-52}{60}\\ \\ \frac{27}{20}=\frac{27\times 3}{20\times 3}=\frac{81}{60}\\ \\ \\ \mathrm{Now},\frac{-42}{60}+\frac{-52}{60}+\frac{81}{60}\\ =\frac{(-42)+(-52)+\left(81\right)}{60}\\ =\frac{-94+81}{60}\\ =\frac{-13}{60}\end{array}$


$\begin{array}{l}\left(iii\right)\\ -1+\frac{7}{-9}+\frac{11}{12}\\ \text{We need a}\text{positive denominator.}\\ \frac{7}{-9}\times \frac{-1}{-1}=\frac{-7}{9}\\ \text{Now},\text{L.C.M. of 1},\text{9 and 12 is 36.}\\ \frac{-1}{1}=\frac{-1\times 36}{1\times 36}=\frac{-36}{36}\\ \\ \frac{-7}{9}=\frac{-7\times 4}{9\times 4}=\frac{-28}{36}\\ \\ \frac{11}{12}=\frac{11\times 3}{12\times 3}=\frac{33}{36}\\ \\ \\ \frac{-36}{36}+\frac{-28}{36}+\frac{33}{36}\\ =\frac{-36-28+33}{36}\\ =\frac{-64+33}{36}\\ =\frac{-31}{36}\\ =\frac{-5}{4}\end{array}$


$\begin{array}{l}\left(iv\right)\\ \frac{-11}{39}+\frac{5}{26}+\frac{2}{1}\\ \text{L.C.M. of 39},\text{26 and 1 is 78.}\\ \\ \frac{-11}{39}=\frac{-11\times 2}{39\times 2}=\frac{-22}{78}\\ \\ \frac{5}{26}=\frac{5\times 3}{26\times 3}=\frac{15}{78}\\ \\ \frac{2}{1}=\frac{2\times 78}{1\times 78}=\frac{156}{78}\\ \text{Now},\frac{-22}{78}+\frac{15}{78}+\frac{156}{78}\\ =\frac{-22+171}{78}\\ =\frac{149}{78}\end{array}$

$$\begin{gathered} \left(v\right) \\ 2+\frac{-1}{2}+\frac{-3}{4} \end{gathered}$$2+−12+−3

$\begin{array}{l}\text{L.C.M. of 2 and 4 is 4.}\\ 2=\frac{2\times 4}{1\times 4}=\frac{8}{4}\\ \frac{-1}{2}=\frac{-1\times 2}{2\times 2}=\frac{-2}{4}\\ \frac{-3}{4}=\frac{-3\times 1}{4\times 1}=\frac{-3}{4}\\ \frac{8}{4}+\frac{(-2)}{4}+\frac{(-3)}{4}\\ =\frac{8-2-3}{4}\\ =\frac{3}{4}\end{array}$

$\begin{array}{l}\left(vi\right)\frac{-9}{11}+\frac{2}{3}+\frac{-3}{4}\\ \text{L.C.M. of 11},\text{3 and 4 is 132.}\\ \frac{-9}{11}=\frac{-9\times 12}{11\times 12}=\frac{-108}{132}\\ \frac{2}{3}=\frac{2\times 44}{3\times 44}=\frac{88}{132}\\ \frac{-3}{4}=\frac{-3\times 33}{4\times 33}=\frac{-99}{132}\\ \frac{-108}{132}+\frac{88}{132}+\frac{(-99)}{132}\\ =\frac{-108+88-99}{132}\\ =\frac{-207+88}{132}=\frac{-119}{132}\end{array}$