2. Verify that $a \div (b + c) \neq (a\div b)+ (a\div c)$ 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

H Harsh Kankaria

$a \div (b + c) \neq (a\div b)+ (a\div c)$

(a) a = 12, b = – 4, c = 2

L.H.S = $\\ a \div (b + c)$

$\\ = 12 \div [(-4)+2] \\ = 12 \div (-2) \\ = - \left (\frac{12}{2} \right ) \\ = -6$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [12\div (-4)]+ (12\div 2) \\ = \left [-\left (\frac{12}{4} \right ) \right ] + \left (\frac{12}{2} \right ) \\ = (-3)+6 \\ = 3$

Therefore. $L.H.S \neq R.H.S$

Hence verified.

(b) a = (–10), b = 1, c = 1

L.H.S = $\\ a \div (b + c)$

$\\ = (-10) \div [1+1] \\ = (-10) \div 2 \\ = - \left (\frac{10}{2} \right ) \\ = -5$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [(-10)\div 1]+ ((-10)\div 1) \\ = \left [-\left (\frac{10}{1} \right ) \right ] + \left [-\left (\frac{10}{1} \right ) \right ] \\ = (-10)+(-10) \\ = -20$

Therefore. $L.H.S \neq R.H.S$

Hence verified.

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