Using the rearrangement property find the sum: (i) 4/3 + 3/5 + (-2)/3 + (-11)/5 (ii) (-8)/3 + (-1)/4 + (-11)/6 + 3/8

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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,

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

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

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

$\frac{-77}{18}$