Question

(11). If a = 27, b = -10 and c = 40
show that a- (b-c) ≠ (a-b) -c .

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Answer 1

Step-by-step explanation:

Given:-

a = 27

b=-10

c= 40

To find:-

Show that a- (b-c) ≠ (a-b) -c

Solution:-

Given that

a = 27

b=-10

c= 40

LHS:-

a- (b-c)

=> 27 -( -10-40)

=> 27 - (-50)

=>27+50

=> 57

a- (b-c) = 57-------(1)

RHS:-

(a-b) -c

=> (27-(-10) ) -40

=> (27+10)-40

=> 37-40

=> -3

(a-b) -c = -3 --------(2)

From (1)&(2)

a- (b-c) ≠ (a-b) -c .

Hence, Proved.

Answer:-

a- (b-c) ≠ (a-b) -c .

Associative Property under subtraction does not holds in the integers.

Used formulae:-

• (+)quantity×(+) quantity= (+)quantity
• (- )quantity×(-)quantity = (+ )quantity
• (+ )quantity×( -) quantity= (-) quantity
• (- )quantity×(+ )quantity=( -)quantity

Answer 2

Answer:

77≠ -3

Hence,not verified

Step-by-step explanation:

If a=27

b= -10

c= 40

Show

a-(-b-c) ≠ (a-b)-c

= 27-{(-10)-(40) ≠{27-(-10)}-40

= 27-(-50)≠ (27+10)-40

= 27+50≠37-40

= 77≠-3

= Hence, not verified.