Question:
Using suitable rearrangement and find the sum:
(a) (4/7) + (-4/9) + (3/7) + (-13/9)
(b) -5 + (7/10) + (3/7) + (-3) + (5/45) + (-4/5)
Solution:
First rearrange the rational numbers and add the numbers with same denominator.
= (4/7) + (3/7) – (4/9) – (13/9)
= ((4 + 3)/7) – ((4 + 13)/9)
= (7/7) – (17/9)
= 1 – (17/9)
= (9 – 17)/9
= -8/9
(b) -5 + (7/10) + (3/7) + (-3) + (5/45) + (-4/5)
= -5 + (-3) + (7/10) + (-4/5) + (3/7) + (5/14)
= 8 + [(7-8)/10] + [(6 + 5)/14]
= – 8 – (1/10) + (11/14)
LCM of 1, 10 and 14 is 70
= (-560 – 7 + 55)/70
= -512/70
= – 256/35