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,
= \((\frac{-3}{1})+(\frac{-23}{18})\)
= \((\frac{-54}{18})+(\frac{-23}{18})\)
= \(\frac{-54+(23)}{18}\)
= \(\frac{-77}{18}\)