Verify that a $\div$ (b $+$ c) ≠ (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
[2 MARKS]

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Solution

Each option: 1 Mark

(a) a = 12, b = −4, c = 2

a $\div$ (b $+$ c) = 12 $\div$ (− 4 $+$ 2) = 12 $\div$ (−2)
= −6

(a $\div$ b) $+$ (a $\div$ c) = [12 $\div$ (−4)] $+$ [12 $\div$ 2] = −3 $+$ 6 = 3

Hence, a $\div$ (b $+$ c) ≠ (a $\div$ b) $+$ (a $\div$ c)

(b) a = −10, b = 1, c = 1

a $\div$ (b $+$ c) = (−10) $\div$ (1 $+$ 1) = (−10) $\div$ 2
= −5

(a $\div$ b) $+$ (a $\div$ c) = [(−10) $\div$ 1] $+$ [(−10) $\div$ 1]
= − 10 − 10 = −20

Hence, a $\div$ (b $+$ c) ≠ (a $\div$ b) $+$ (a $\div$ c)

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