Question
Complete the following table by finding the sums.
✓
Answer
Let
Here,
$a. =\frac{2}{3}+\Big(-\frac{1}{9}\Big)$
$=\frac{2}{3}-\frac{1}{9}$
$=\frac{2\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{6}{9}-\frac{1}{9}$
$=\frac{6-1}{9}$
$=\frac{5}{9}$
$b. =\frac{2}{3}+\frac{4}{11}$
$=\frac{2\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{22}{33}+\frac{12}{33}$
$=\frac{22+12}{33}$
$=\frac{34}{33}$
$c. =\frac{2}{3}+\Big(-\frac{5}{6}\Big )$
$=\frac{2}{3}-\frac{5}{6}$
$=\frac{2\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{4}{6}-\frac{5}{6}$
$=\frac{4-5}{6}$
$=-\frac{1}{6}$
$d. =-\frac{5}{4}+\Big(-\frac{1}{9}\Big) $
$=\frac{-5}{4}-\frac{1}{9}$
$=\frac{-5\times9}{4\times9}-\frac{1\times4}{9\times4}$
$=\frac{-45}{36}-\frac{4}{36}$
$=\frac{-45-4}{36}$
$=\frac{-49}{36}$
$e. =-\frac{5}{4}+\Big(-\frac{5}{6}\Big)$
$=-\frac{5}{4}-\frac{5}{6}$
$=\frac{-5\times3}{4\times3}-\frac{5\times2}{6\times2}$
$=\frac{-15}{12}-\frac{10}{12}$
$=\frac{-15-10}{12}$
$=-\frac{25}{12}$
$f. =-\frac{1}{3}+\Big(-\frac{1}{9}\Big)$
$=-\frac{1}{3}-\frac{1}{9}$
$=\frac{-1\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{-3}{9}-\frac{1}{9}$
$=\frac{-3-1}{9}$
$=\frac{-4}{9}$
$g. =-\frac{1}{3}+\frac{4}{11}$
$=\frac{-1\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{-11}{33}+\frac{12}{33}$
$h. =-\frac{1}{3}+\Big(\frac{-5}{6}\Big)$
$=-\frac{1}{3}-\frac{5}{6}$
$=\frac{-1\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{-2}{6}-\frac{5}{6}$
$=\frac{-2-5}{6}$
$=\frac{-7}{6}$
Hence, the complete table is:
Here,
$a. =\frac{2}{3}+\Big(-\frac{1}{9}\Big)$
$=\frac{2}{3}-\frac{1}{9}$
$=\frac{2\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{6}{9}-\frac{1}{9}$
$=\frac{6-1}{9}$
$=\frac{5}{9}$
$b. =\frac{2}{3}+\frac{4}{11}$
$=\frac{2\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{22}{33}+\frac{12}{33}$
$=\frac{22+12}{33}$
$=\frac{34}{33}$
$c. =\frac{2}{3}+\Big(-\frac{5}{6}\Big )$
$=\frac{2}{3}-\frac{5}{6}$
$=\frac{2\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{4}{6}-\frac{5}{6}$
$=\frac{4-5}{6}$
$=-\frac{1}{6}$
$d. =-\frac{5}{4}+\Big(-\frac{1}{9}\Big) $
$=\frac{-5}{4}-\frac{1}{9}$
$=\frac{-5\times9}{4\times9}-\frac{1\times4}{9\times4}$
$=\frac{-45}{36}-\frac{4}{36}$
$=\frac{-45-4}{36}$
$=\frac{-49}{36}$
$e. =-\frac{5}{4}+\Big(-\frac{5}{6}\Big)$
$=-\frac{5}{4}-\frac{5}{6}$
$=\frac{-5\times3}{4\times3}-\frac{5\times2}{6\times2}$
$=\frac{-15}{12}-\frac{10}{12}$
$=\frac{-15-10}{12}$
$=-\frac{25}{12}$
$f. =-\frac{1}{3}+\Big(-\frac{1}{9}\Big)$
$=-\frac{1}{3}-\frac{1}{9}$
$=\frac{-1\times3}{3\times3}-\frac{1\times1}{9\times1}$
$=\frac{-3}{9}-\frac{1}{9}$
$=\frac{-3-1}{9}$
$=\frac{-4}{9}$
$g. =-\frac{1}{3}+\frac{4}{11}$
$=\frac{-1\times11}{3\times11}+\frac{4\times3}{11\times3}$
$=\frac{-11}{33}+\frac{12}{33}$
$h. =-\frac{1}{3}+\Big(\frac{-5}{6}\Big)$
$=-\frac{1}{3}-\frac{5}{6}$
$=\frac{-1\times2}{3\times2}-\frac{5\times1}{6\times1}$
$=\frac{-2}{6}-\frac{5}{6}$
$=\frac{-2-5}{6}$
$=\frac{-7}{6}$
Hence, the complete table is:
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