Question 87

Complete the following table by finding the sums.

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Solution

Let the missing values be a, b, c, d, e, f, g, and h respectively in the mentioned order.

Such that a is the sum of respective values of row and column number.

Here, $a = \frac{2}{3} + (- \frac{1}{9}) = \frac{2}{3} - \frac{1}{9} = \frac{2 \times 3}{3 \times 3} - \frac{1 \times 1}{9 \times 1} = \frac{6}{9} - \frac{1}{9} = \frac{6 - 1}{9} = \frac{5}{9}$
$b = \frac{2}{3} + \frac{4}{11} = \frac{2 \times 11}{3 \times 11} + \frac{4 \times 3}{11 \times 3} = \frac{22}{33} + \frac{12}{33} = \frac{22 + 12}{33} = \frac{34}{33}$
$c = \frac{2}{3} + (- \frac{5}{6}) = \frac{2}{3} - \frac{5}{6} = \frac{2 \times 2}{3 \times 3} - \frac{5 \times 1}{6 \times 1} = \frac{4}{6} - \frac{5}{6} = \frac{4 - 5}{6} = - \frac{1}{6}$
$d = - \frac{5}{4} + (- \frac{1}{9}) = \frac{- 5}{4} - \frac{1}{9} = \frac{- 5 \times 9}{4 \times 9} - \frac{1 \times 4}{9 \times 4} = \frac{- 45}{36} - \frac{4}{36} = \frac{- 45 - 4}{36} = \frac{- 49}{36}$
$e = - \frac{5}{4} + (- \frac{5}{6}) = - \frac{5}{4} - \frac{5}{6} = \frac{- 5 \times 3}{4 \times 3} - \frac{5 \times 2}{6 \times 2} = \frac{- 15}{12} - \frac{10}{12} = \frac{- 15 - 10}{12} = - \frac{25}{12}$
$f = - \frac{1}{3} + (- \frac{1}{9}) = - \frac{1}{3} - \frac{1}{9} = \frac{- 1 \times 3}{3 \times 3} - \frac{1 \times 1}{9 \times 1} = \frac{- 3}{9} - \frac{1}{9} = \frac{- 3 - 1}{9} = \frac{- 4}{9}$
$g = - \frac{1}{3} + \frac{4}{11} = \frac{- 1 \times 11}{3 \times 11} + \frac{4 \times 3}{11 \times 3} = \frac{- 11}{33} + \frac{12}{33} = \frac{- 11 + 12}{33} = \frac{1}{33}$
$h = - \frac{1}{3} + (\frac{- 5}{6}) = - \frac{1}{3} - \frac{5}{6} = \frac{- 1 \times 2}{3 \times 2} - \frac{5 \times 1}{6 \times 1} = \frac{- 2}{6} - \frac{5}{6} = \frac{- 2 - 5}{6} = \frac{- 7}{6}$

Hence, the complete table is

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Question 87 Complete the following table by finding the sums.

Question 87

Complete the following table by finding the sums.