Question

subtract a/6+b/12+c/20 from the sum of a/2+b/3+c/4and 2a/3+3b/4+4c/5​

Answer 1

Answer:

a+b+c

Step-by-step explanation:

(a/2+b/3+c/4)+(2a/3+3b/4+4c/5)

after taking LCM of denominators

=(6a+4b+3c/12)+(40a+45b+48c/60)

=(5(6a+4b+3c)+(40a+45b+48c))/60

=(30a+20b+15c+40a+45b+48c)/60

=(70a+65b+63c)/60

(70a+65b+63c)/60-(a/6+b/12+c/20)

=(70a+65b+63c)/60-(10a+5b+3c)/60

=(70a+65b+63c-10a-5b-3c)/60

=(60a+60b+60c)/60

=60(a+b+c)/60

=(a+b+c)