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Using the rearrangement property find the sum:

(i) \(\frac{4}{3}+\frac{3}{5}+\frac{-2}{3}+\frac{-11}{5}\) 

(ii) \(\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}\)

(iii) \(\frac{-13}{20}+\frac{11}{14}+\frac{-5}{7}+\frac{3}{10}\)

(iv) \(\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}\)

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Rearrangement property says that, the numbers in an addition expression may be arranged and grouped in any order. 

Therefore,

(i) \(\frac{4}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-11}{5}\)

We arrange the numbers with same denominators together,

\((\frac{4}{3}+\frac{-2}{3})\) + \((\frac{3}{5}+\frac{-11}{5})\)

\((\frac{4+(-2)}{3}) +(\frac{3+(-11)}{5})\)

\((\frac{2}{3})+(\frac{-8}{5})\)

 Now, we take LCM of 3 and 5=15

\(\frac{2}{3}= \frac{2\times3}{5\times3} =\frac{10}{15}\)

And,

\(\frac{-8}{5}= \frac{-8\times3}{5\times3} =\frac{-24}{15}\)

Therefore,

\((\frac{2}{3})\) + \((\frac{-8}{5})\)

\((\frac{10}{15})+(\frac{-8}{5})\)

\(\frac{10+(-24)}{15}\)

\(\frac{-14}{15}\)

(ii)

\(\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}\)

We arrange the numbers,

=  \((\frac{-8}{3}+\frac{-11}{6})\) + \((\frac{-1}{4}+\frac{3}{8})\) 

LCM of 3 and 6 =6

\(\frac{-8}{3}= \frac{-8\times2}{3\times2} =\frac{-16}{6}\)

And,

\(\frac{-11}{6}= \frac{-11\times1}{6\times1} =\frac{-11}{6}\)

LCM of 4 and 8 =8

\(\frac{-1}{4}= \frac{-1\times2}{4\times2} =\frac{-2}{8}\)

And,

\(\frac{3}{8}= \frac{3\times1}{8\times1} =\frac{3}{8}\)

Now, 

\((\frac{-16}{6}+\frac{-11}{6})\) + \((\frac{-2}{8}+\frac{3}{8})\)

\((\frac{-16+(-11)}{6})+(\frac{-2+3}{8})\)

\((\frac{-27}{6})+(\frac{1}{8})\)

Now, we take LCM of 6 and 8=24

\(\frac{-27}{6}= \frac{-27\times4}{6\times4}= \frac{-108}{24}\)

And,

\(\frac{1}{8}= \frac{-1\times3}{8\times3} = \frac{3}{24}\)

Therefore,

\((\frac{-27}{6})+(\frac{1}{8})\)

\((\frac{-108}{24})+(\frac{3}{24})\)

\(\frac{-108+3}{24}\)

\(\frac{-105}{24}\)

In lowest terms

\(\frac{-105}{24}= \frac{-105÷3}{24÷3} =\frac{-35}{8}\)

(iii) 

\(\frac{-13}{20}+\frac{11}{14}+\frac{-5}{7}+\frac{7}{10}\)

We arrange the numbers,

\((\frac{-13}{20}+\frac{7}{10})\) + \((\frac{11}{14}+\frac{-5}{7})\)

LCM of 20 and 10 =20

\(\frac{-13}{20}= \frac{-13\times1}{20\times1} = \frac{-13}{20}\)

And,

\(\frac{7}{10}=\frac{7\times2}{10\times1}= \frac{14}{20}\)

LCM of 14 and 7 =14

\(\frac{11}{14}=\frac{11\times1}{14\times1}=\frac{11}{14}\)

And,

\(\frac{-5}{7}= \frac{-5\times2}{7\times2}=\frac{-10}{14}\)

Now,

\((\frac{-13}{20}+\frac{14}{20})+(\frac{11}{14}+\frac{-10}{14})\)

\((\frac{-13+14}{20})+(\frac{11+(-10)}{14})\) 

\((\frac{1}{20})+(\frac{1}{14})\)

Now, we take LCM of 20 and 14=140

\(\frac{1}{20}=\frac{1\times7}{20\times7}=\frac{7}{140}\)

And,

\(\frac{1}{14}=\frac{1\times10}{14\times10}=\frac{10}{140}\)

Therefore,

\((\frac{1}{20}) +(\frac{1}{14})\)

\((\frac{7}{140})+(\frac{10}{140})\)

\(\frac{7+10}{140}\) 

\(\frac{17}{140}\)

(iv) 

\(\frac{-6}{7}+\frac{-5}{6}\frac{-4}{9}+\frac{-15}{7}\)

We arrange the numbers,

\((\frac{-6}{7}+\frac{-15}{7})+(\frac{-5}{6}+\frac{-4}{9})\)

LCM of 4 and 9 =18

And,

\(\frac{-5}{6}=\frac{-4\times2}{9\times4}= \frac{-8}{18}\)

Now,

\((\frac{-6}{7}+\frac{-15}{7})+(\frac{-8}{18}+\frac{-15}{18})\)

\((\frac{-6+(-15)}{7})+(\frac{-8+(-15)}{18})\)

\((\frac{-6+(-15)}{7})+(\frac{-8-15}{18})\)

\((\frac{-21}{7})+(\frac{-23}{18})\)

In lowest terms,

\(\frac{-21}{7}=\frac{-21÷7}{7÷7}= \frac{-3}{1}\)

Now, we take LCM of 1 and 18=18

\(\frac{-3}{1}= \frac{-3\times18}{1\times18} =\frac{-54}{18}\)

And,

\(\frac{-23}{18}= \frac{-23\times1}{18\times1} =\frac{-23}{18}\)

Therefore,

\((\frac{-3}{1})+(\frac{-23}{18})\)

\((\frac{-54}{18})+(\frac{-23}{18})\)

\(\frac{-54+(23)}{18}\)

\(\frac{-77}{18}\) 

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