Question

Verify the property: 1 (a×b)×c =a×(b×c).

2 a×(b+c)=(a×b)+(a×c)

By taking a= -2/7 b= -5/6 & c = 1/4

Answer 1

Given that to verify the properties :

• (a × b) × c = a × (b × c)
• a × (b + c) = (a × b) + (a × c)
• When, a = -2/7 , b = -5/6 & c = 1/4

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$\underline{\underline{\large{\bold\blue{Verifying \: property \: 1 \: : }}}}$

$\longrightarrow\sf (a \times b)\times c = a \times (b \times c) \\\\\qquad\small\sf {\dag \: Putting \: values \: of \: a,b\: \& \: c } \\\\\longrightarrow\sf \bigg(\dfrac{-2}{7} \times \dfrac{-5}{6} \bigg )\times \dfrac{1}{4} = \dfrac{-2}{7}\times \bigg(\dfrac{-5}{6}\times \dfrac{1}{4} \bigg) \\\\\longrightarrow\sf \bigg(\dfrac{\cancel{10}}{\cancel{42}} \bigg)\times \dfrac{1}{4} = \dfrac{\cancel{-2}}{7}\times \bigg(\dfrac{-5}{\cancel{24}} \bigg)\\\\\longrightarrow\sf \dfrac{5}{21} \times \dfrac{1}{4} = \dfrac{1}{7} \times \dfrac{\cancel{-}5}{\cancel{-}12} \\\\\longrightarrow\large\sf\blue{\dfrac{5}{84} = \dfrac{5}{84}}$

Hence,verified!

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$\underline{\underline{\large{\bold\red{Verifying \: property \: 2 \: : }}}}$

$\longrightarrow\sf a\times (b + c) = (a \times b) + (a \times c) \\\\\qquad\small\sf {\dag \: Putting\: values \: of \: a,b \: \& \: c }\\\\\longrightarrow\sf \dfrac{-2}{7} \times \bigg(\dfrac{-5}{6} + \dfrac{1}{4} \bigg) = \bigg(\dfrac{\cancel{-2}}{7} \times \dfrac{-5}{\cancel{6}} \bigg) + \bigg(\dfrac{\cancel{-2}}{7}\times \dfrac{1}{\cancel{4}} \bigg) \\\\\longrightarrow\sf \dfrac{-2}{7}\times \bigg(\dfrac{-20 + 6}{24} \bigg) = \bigg(\dfrac{5}{21} \bigg) + \bigg(\dfrac{-1}{14} \bigg) \\\\\longrightarrow\sf \cancel{\dfrac{-2}{7}} \times \cancel{\dfrac{-14}{24}} = \bigg(\dfrac{20 - 6}{84} \bigg) \\\\\longrightarrow\sf \dfrac{1}{6} = \cancel{ \bigg(\dfrac{14}{84}\bigg)} \\\\\longrightarrow\large\sf\red{\dfrac{1}{6} = \dfrac{1}{6}}$

Hence, verified!

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Answer 2

$\huge{\boxed{\underline{\overline{\pink{Solution}}}}}$

$\huge{\boxed{\mathsf{\green{GIVEN}}}}$

→ A = $\frac{- 2 }{7}$

→ B = $\frac{- 5}{6}$

→ C = $\frac{ 1 }{4}$

$\huge{\boxed{\mathsf{\green{To\: Prove}}}}$

1 . ( A × B ) × C = A × ( B × C )

2. A × ( B + C ) = ( A × B ) + ( A × B )

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First Case -

Taking LHS of the first property :-

→ ( A × B ) × C = $( \frac{-2}{7} \times \frac{-5}{6} ) \times \frac{1}{4}$

→ $( \frac{10}{42}) \times \frac{1}{4}$

→ $\frac{5}{42 \times 2 }$

→ $\frac{5 }{84}$

Taking RHS of the first property :-

→ A × ( B × C ) = $\frac{-2}{7} \times (\frac{-5}{6} \times \frac{1}{4} )$

→ $\frac{-2}{7} \times ( \frac{-5}{24} )$

→ $\frac{1}{7} \times \frac{5}{12}$

→ $\frac{5}{84}$

LHS = RHS

Hence proved .

Second case :-

Taking LHS of the second property :-

→ A×(B+C) = $\frac{-2}{7} \times ( \frac{-5}{6} + \frac{1}{4})$

→ $\frac{-2}{7} \times \frac{-10 + 3 }{12}$

→ $\frac{-2}{7}\times \frac{-7}{12}$

→ $\frac{ 1 }{6}$

Taking RHS of the second property :-

→ (A × B ) + ( A× C) = $\frac{-2\times-5}{7\times6} + \frac{-2\times1}{7 \times 4}$

→ $\frac{5}{21} + \frac{-1}{14}$

→ $\frac{10 - 3}{42}$

→ $\frac{7 }{42}$

→ $\frac{1 }{6}$

LHS = RHS

Hence proved .