book : simple maple, subjet: math, class: 8, lesson: 1, exercise: 1.1
Answers
Answer:
1. Using appropriate properties find.
(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6
Solution:
-2/3 × 3/5 + 5/2 – 3/5 × 1/6
= -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity)
= 3/5 (-2/3 – 1/6)+ 5/2
= 3/5 ((- 4 – 1)/6)+ 5/2
= 3/5 ((–5)/6)+ 5/2 (by distributivity)
= – 15 /30 + 5/2
= – 1 /2 + 5/2
= 4/2
= 2
(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
ans=Solution:
2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5
= 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)
= 2/5 × (- 3/7 + 1/14) – 3/12
= 2/5 × ((- 6 + 1)/14) – 3/12
= 2/5 × ((- 5)/14)) – 1/4
= (-10/70) – 1/4
= – 1/7 – 1/4
= (– 4– 7)/28
= – 11/28
2Write the additive inverse of each of the following
ans=Solution:
(i) 2/8
Additive inverse of 2/8 is – 2/8
(ii) -5/9
Additive inverse of -5/9 is 5/9
(iii) -6/-5 = 6/5
Additive inverse of 6/5 is -6/5
(iv) 2/-9 = -2/9
Additive inverse of -2/9 is 2/9
(v) 19/-16 = -19/16
Additive inverse of -19/16 is 19/16
3 Verify that: -(-x) = x for.
(i) x = 11/15
(ii) x = -13/17
ans=Solution:
(i) x = 11/15
We have, x = 11/15
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of 11/15 is – 11/15 (as 11/15 + (-11/15) = 0)
The same equality 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.
Or, – (-11/15) = 11/15
i.e., -(-x) = x
(ii) -13/17
We have, x = -13/17
The additive inverse of x is – x (as x + (-x) = 0)
Then, the additive inverse of -13/17 is 13/17 (as 11/15 + (-11/15) = 0)
The same equality (-13/-13/17 + 13/17) = 0, shows that the additive inverse of 13/17 is -13/17.
Or, – (13/17) = -13/17,
i.e., -(-x) = x