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Answer 1

We're asked to verify that (x + y) + z = x + (y + z) for two sets of values for 'x', 'y', and 'z'.

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Part (I):

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$\sf \dashrightarrow \ x = \dfrac{1}{2}$

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$\sf \dashrightarrow \ y = \dfrac{2}{3}$

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$\sf \dashrightarrow \ z = -\dfrac{1}{5}$

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We're asked to prove that;

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$\sf \dashrightarrow (x + y) + z = x + (y + z)$

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Now we'll substitute the values given in the question.

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$\sf \dashrightarrow \Bigg[\dfrac{1}{2} + \dfrac{2}{3}\Bigg] - \dfrac{1}{5} = \dfrac{1}{2} + \Bigg[\dfrac{2}{3} - \dfrac{1}{5}\Bigg]$

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Taking LCM;

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$\sf \dashrightarrow \Bigg[\bigg(\dfrac{1}{2} \times \dfrac{3}{3}\bigg) + \bigg(\dfrac{2}{3} \times \dfrac{2}{2}\bigg)\Bigg] - \dfrac{1}{5} = \dfrac{1}{2} + \Bigg[\bigg(\dfrac{2}{3} \times \dfrac{5}{5} \bigg) - \bigg(\dfrac{1}{5} \times \dfrac{3}{3} \bigg)\Bigg]$

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$\sf \dashrightarrow \Bigg[\dfrac{3}{6} + \dfrac{4}{6}\Bigg] - \dfrac{1}{5} = \dfrac{1}{2} + \Bigg[\dfrac{10}{15} - \dfrac{3}{15}\Bigg]$

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$\sf \dashrightarrow \Bigg[\dfrac{3 + 4}{6}\Bigg] - \dfrac{1}{5} = \dfrac{1}{2} + \Bigg[\dfrac{10 - 3}{15}\Bigg]$

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$\sf \dashrightarrow \dfrac{7}{6} - \dfrac{1}{5} = \dfrac{1}{2} + \dfrac{7}{15}$

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On taking LCM again we get;

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$\sf \dashrightarrow \Bigg(\dfrac{7}{6} \times \dfrac{5}{5}\Bigg) - \Bigg(\dfrac{1}{5} \times \dfrac{6}{6}\Bigg) = \Bigg(\dfrac{1}{2} \times \dfrac{15}{15} \Bigg) +\Bigg(\dfrac{7}{15} \times \dfrac{2}{2} \Bigg)$

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$\sf \dashrightarrow \dfrac{35}{30} - \dfrac{6}{30} = \dfrac{15}{30} + \dfrac{14}{30}$

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$\sf \dashrightarrow \dfrac{35 - 6}{30} = \dfrac{15 + 14}{30}$

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$\sf \dashrightarrow \dfrac{29}{30} = \dfrac{29}{30}$

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LHS = RHS

Hence proved.

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Part(II)

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$\sf \dashrightarrow \ x = \dfrac{-2}{5}$

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$\sf \dashrightarrow \ y = \dfrac{4}{3}$

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$\sf \dashrightarrow \ z = \dfrac{-7}{10}$

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We're asked to prove that;

‎

$\sf \dashrightarrow (x + y) + z = x + (y + z)$

Now we'll substitute the values;

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$\sf \dashrightarrow \Bigg[\dfrac{-2}{5} + \dfrac{4}{3}\Bigg] + \dfrac{-7}{10} = \dfrac{-2}{5} + \Bigg[\dfrac{4}{3} + \dfrac{-7}{10}\Bigg]$

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On taking LCM we get;

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$\sf \dashrightarrow \Bigg[\bigg(\dfrac{-2}{5} \times \dfrac{3}{3}\bigg) + \bigg(\dfrac{4}{3} \times \dfrac{5}{5}\bigg)\Bigg] - \dfrac{7}{10} = \dfrac{-2}{5} + \Bigg[\bigg(\dfrac{4}{3} \times \dfrac{10}{10} \bigg) + \bigg(\dfrac{-7}{10} \times \dfrac{3}{3} \bigg)\Bigg]$

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$\sf \dashrightarrow \Bigg[\dfrac{-6}{15} + \dfrac{20}{15}\Bigg] - \dfrac{7}{10} = \dfrac{-2}{5} + \Bigg[\dfrac{40}{30} + \bigg(\dfrac{-21}{30}\bigg)\Bigg]$

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$\sf \dashrightarrow \Bigg[\dfrac{-6 + 20}{15}\Bigg] - \dfrac{7}{10} = \dfrac{-2}{5} + \Bigg[\dfrac{40 - 21}{30}\Bigg]$

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$\sf \dashrightarrow \Bigg[\dfrac{14}{15}\Bigg] - \dfrac{7}{10} = \dfrac{-2}{5} + \Bigg[\dfrac{19}{30}\Bigg]$

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On taking LCM again we get;

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$\sf \dashrightarrow \bigg(\dfrac{14}{15} \times \dfrac{2}{2}\bigg) - \bigg(\dfrac{7}{10} \times \dfrac{3}{3} \bigg) = \bigg(\dfrac{-2}{5} \times \dfrac{6}{6}\bigg) + \dfrac{19}{30}$

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$\sf \dashrightarrow \dfrac{28}{30} - \dfrac{21}{30} = \dfrac{-12}{30} + \dfrac{19}{30}$

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$\sf \dashrightarrow \dfrac{28 - 21}{30} = \dfrac{-12 + 19}{30}$

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$\sf \dashrightarrow \dfrac{7}{30} = \dfrac{7}{30}$

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LHS = RHS

Hence proved.

Answer 2

1]

We have

x = ½

y = ⅔

z = -⅕

So,

LHS

$\sf \bigg(\dfrac{1}{2} + \dfrac{2}{3} \bigg) - \dfrac{1}{5}$

$\sf \bigg(\dfrac{3 + 4}{6}\bigg) - \dfrac{1}{5}$

$\sf \dfrac{7}{6} - \dfrac{1}{5}$

$\sf \dfrac{35 - 6}{30}$

$\sf \dfrac{29}{30}$

RHS

$\sf \dfrac{1}{2} + \bigg( \dfrac{2}{3} - \dfrac{1}{5}\bigg)$

$\sf \dfrac{1}{2} + \dfrac{10 - 3}{15}$

$\sf \dfrac{1}{2} + \dfrac{7}{15}$

$\sf \dfrac{15 + 14}{30}$

$\sf \dfrac{29}{30}$

Hence, proved

2]

LHS

$\sf \bigg( \dfrac{-2}{5} + \dfrac{4}{3}\bigg) - \dfrac{7}{10}$

$\sf \bigg(\dfrac{-6 + 20}{15}\bigg) \dfrac{7}{10}$

$\sf \dfrac{14}{15} - \dfrac{7}{10}$

$\sf \dfrac{28 - 21}{30}$

$\sf \dfrac{7}{30}$

RHS

$\sf \dfrac{-2}{5}+ \bigg(\dfrac{4}{3}+ \dfrac{-7}{10}\bigg)$

$\sf \dfrac{-2}{5}+ \bigg(\dfrac{4}{3}-\dfrac{7}{10}\bigg)$

$\sf \dfrac{-2}{5} + \bigg(\dfrac{40 - 21}{30}\bigg)$

$\sf \dfrac{-2}{5} + \dfrac{19}{30}$

$\sf \dfrac{-12+19}{30}$

$\sf \dfrac{7}{30}$

Hence, proved