# Using the rearrangement property find the sum:

Question:

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

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

Solution:

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

$=\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)$

$=\left(\frac{2}{3}+\frac{-8}{5}\right)$

$=\left(\frac{10}{15}+\frac{-24}{15}\right)$

$=\left(\frac{10-24}{15}\right)$

$=\frac{-14}{15}$

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

$=\left(\frac{-16}{6}+\frac{-11}{6}\right)+\left(\frac{-2}{8}+\frac{3}{8}\right)$

$=\left(\frac{-16-11}{6}\right)+\left(\frac{-2+3}{8}\right)$

$=\left(\frac{-27}{6}+\frac{1}{8}\right)$

$=\left(\frac{-108}{24}+\frac{3}{24}\right)$

$=\frac{-105}{24}$

$=\frac{35}{8}$

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

$=\left(\frac{-13}{20}+\frac{14}{20}\right)+\left(\frac{11}{14}+\frac{-10}{14}\right)$

$=\left(\frac{-13+14}{20}\right)+\left(\frac{11-10}{14}\right)$

$=\left(\frac{1}{20}+\frac{1}{14}\right)$

$=\left(\frac{7}{140}+\frac{10}{140}\right)$

$=\left(\frac{7+10}{140}\right)$

$=\left(\frac{17}{140}\right)$

$=\frac{17}{140}$

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

$=\left(\frac{-6}{7}+\frac{-15}{7}\right)+\left(\frac{-15}{18}+\frac{-8}{18}\right)$

$=\left(\frac{-6-15}{7}\right)+\left(\frac{-15-8}{18}\right)$

$=\left(\frac{-21}{7}+\frac{-23}{18}\right)$

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

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

$=\left(\frac{-54-23}{18}\right)$

$=\frac{-77}{18}$