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Subtraction of Rational Numbers

k Taking x = ...

Question

Question 94 (k)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18},$ find
$x - (y + z)$

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Solution

Given, $x = \frac{- 4}{9}, y = \frac{5}{12} a n d z = \frac{7}{18}$
Here, $x - (y + z) = \frac{- 4}{9} - (\frac{5}{12} + \frac{7}{18}) = \frac{- 4}{9} - (\frac{5 \times 3 + 7 \times 2}{36}) = \frac{- 4}{9} - (\frac{15 + 14}{36})$ [ $∵$ LCM of 12 and 18 = 36]
$= \frac{- 4}{9} - \frac{29}{36} = \frac{- 4 \times 4 - 29}{36} = \frac{- 16 - 29}{36}$ [ $∵$ LCM of 9 and 36 = 36]
$= \frac{- 45}{36} = \frac{- 5}{4}$

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Similar questions

Question 94 (i)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18},$

Find $x + (y + z)$ .

Question 94 (h)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18}$

Find $(x - y) + z$ .

Question 94 (j)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18}$ .

Find $x \div (y \div z)$ .

Question 94 (g)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18}$

Find $(x \div y) \times z$ .

Question 94 (e)

Taking $x = \frac{- 4}{9}, y = \frac{5}{12}$ and $z = \frac{7}{18},$ find
The reciprocal of x + y.

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Subtraction of Rational Numbers

Standard VII Mathematics

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