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,

\[a\div \left( b+c \right)\ne \left( a\div b \right)+\left( a\div c \right)\dots\left( 1 \right)\]

a) \(a = 12, b = –4, c = 2\)

Take L.H.S, \(a \div(b+c)\)

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

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

Now take R.H.S

\[\begin{align}&(a \div b)+(a \div c) \\ &\left[ 12\div (-4) \right]+\left( 12\div 2 \right) \\ \ \ \ &=\left[ 12\times \frac{(-1)}{4} \right]+\left( 12\times \frac{1}{2} \right) \\ &=\frac{\left( -12 \right)}{4}+\left( \frac{12}{2} \right) \\ & =-3+6 \\ & =3 \\ \end{align}\]

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

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

Hence verified.

b) \(a =(-10),\;b=1,\;c=1\)

Take L.H.S, \(a (b + c)\)

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

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

Take R.H.S,\((a \div b)+(a \div c)\)

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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