Ex.1.4 Q2 Integers Solution - NCERT Maths Class 7

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Question

Verify that \(a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)\) for each of the following values of \(a, b\) and \(c.\)

(a) \( a=12,b=-4,c=2\)

(b) \(a=\left( -10 \right),\,\,b=1,\,\,c=1\)

Text Solution

   

Steps:

Let,

\[a\div \left( b\text{ }+\text{ }c \right)\ne \left( a\text{ }\div \text{ }b \right)+\left( a\text{ }\div \text{ }c \right)~~~~~~~~~~~~~~~~~~\left( 1 \right)\]

a) \(a = 12, b = –4, c = 2\)

Take L.H.S, \(\text{a}\,\text{ }\!\!\div\!\!\text{ }\,\left( \text{b}\,\text{+}\,\text{c} \right)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align}   \text{a}&\div (\text{b}+\text{c})  \\   &=12\div \left( -4+2 \right)  \\   &=12\div \left( -2 \right)  \\   &=12\times \frac{1}{\left( -2 \right)}  \\  & =\frac{12}{\left( -2 \right)} \\  & =-6 \\ \end{align}\]

Now take R.H.S, \[\begin{align}   \left( \text{a}\div \text{b} \right)&+\left( \text{a}\div \text{c} \right)  \\   \left[ 12\div (-4) \right]&+\left( 12\div 2 \right)  \\   \ \ \ &=\left[ 12\times \frac{(-1)}{4} \right]+\left( 12\times \frac{1}{2} \right)  \\   &=\frac{\left( -12 \right)}{4}+\left( \frac{12}{2} \right)  \\  & =-3+6 \\  & =3 \\ \end{align}\]

Putting the values of L.H.S and R.H.S in equation (\(1\)), we get

\[\begin{align}  \text{a }\left( \text{b+c} \right)&\ne \left( \text{a}\,\text{ }\text{ }\,\text{b} \right)\text{+}\left( \text{a}\,\text{ }\text{ }\,\text{c} \right)  \\   6&\ne 3  \\   \text{ L}\text{.H}\text{.S }&\ne \text{ R}\text{.H}\text{.S }  \\\end{align}\]

Hence verified.

b) \(\text{a = }\left( \text{-10} \right)\text{, b=1, c=1}\)

Take L.H.S, \(a \left( \text{b + c} \right)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align}   &=\text{(}-\text{10)}\,\text{ }\text{ }\,\text{(1+1)}  \\    &=\text{(}-\text{10)}\,\text{ }\text{ }\,\text{(2)}  \\   &=\text{(}-\text{10)}\,\text{ }\!\!\times\!\!\text{ }\,\frac{\text{1}}{\text{2}}  \\   &=-\text{5}  \\\end{align}\]

Take R.H.S, \(\left( \text{a }\text{b} \right)\text{ + }\left( \text{a  }\text{  c} \right)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align}&= [(–10) ÷1] + [(–10) ÷1]\\ &= [(–10) × ] + [(–10) × ]\\ &= (–10) + (–10)\\ &= – 10 –10\\ &= –20\end{align}\]

Putting the values of L.H.S and R.H.S in equation (\(1\)), we get

\[\begin{align} a ÷ (b + c) &≠ (a ÷ b) + (a ÷ c)\\ –5& ≠ –20\end{align}\]

L.H.S \(≠\) R.H.S

 Hence verified

  
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