# Verify that $a Ă· (b+c) â‰ (a Ă· b) + (a Ă· c)$ for each of the following values of $a,\ b$ and $c$.(a) $a=12,\ b=- 4,\ c=2$(b) $a=(-10),\ b = 1,\ c = 1$

To do:

We have to verify that $a \div (b+c) ≠ (a \div b) + (a \div c)$ for each of the given values of $a,\ b$ and $c$.

Solution:

We know that,

$a \div b=a \times \frac{1}{b}$

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

Here,

L.H.S.$=a\div (b+c)$

$=12\div (-4+2)$          [On substituting the values of $a,\ b$ and $c$]

$=12\div (-2)$

$=12\times\frac{1}{-2}$

$=-6$

R.H.S.$=(a\div b)+(a\div c)$

$=(12\div -4)+(12\div 2)$

$=(12\times\frac{1}{-4})+(12\times\frac{1}{2})$

$=(-3)+(6)$

$=3$

Here, we can see that, L.H.S.$≠$ R.H.S.

Hence verified.

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

L.H.S.$=a\div (b+c)$

$=-10\div (1+1)$

$=-10\div 2$

$=-10\times\frac{1}{2}$

$=-5$

R.H.S.$=(a\div b)+(a\div c)$

$=(-10\div 1)+(-10\div 1)$

$=(-10\times\frac{1}{1})+(-10\times\frac{1}{1})$

$=(-10)+(-10)$

$=-20$

Here, L.H.S. $≠$ R.H.S.

Hence verified.

Updated on: 10-Oct-2022

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