Question Number 179140 by Shrinava last updated on 25/Oct/22

(1/(3a)) = (1/(4b)) = (1/(6c)) and a+b+c=27 find a−c=?

Answered by Frix last updated on 25/Oct/22

a=4t b=3t c=2t a+b+c=27 9t=27 t=3 a=12 b=9 c=6 a−c=6

Commented byRasheed.Sindhi last updated on 25/Oct/22

Typo sir: a+b+c=27⇒9t=27

Commented byFrix last updated on 25/Oct/22

yes thank you

Answered by Rasheed.Sindhi last updated on 25/Oct/22

3a=4b=6c ∧ a+b+c=27; a−c=? [ _( _( ) ) 3a=6c⇒a−2c=0⇒a−c=c_(Hence we need value of c) ] ▶Eleminating b from a+b+c=27 3a=4b=6c⇒ { ((b=(3/2)c)),((b=(3/4)a)) :} •a+b+c=27⇒a+(3/4)a+c=27 ⇒7a+4c=108....(i) •a+b+c=27⇒a+(3/2)c+c=27 ⇒2a+5c=54....(ii) ▶Eleminating a from (i) & (ii) { ((2×(i)⇒14a+8c=216)),((7×(ii)⇒14a+35c=378)) :}⇒27c=162 ⇒c=((162)/(27))=6 •3a=6c⇒a−2c=0⇒a−c=c=6

Answered by som(math1967) last updated on 25/Oct/22

(1/(3a))=(1/(4b))=(1/(6c)) 3a=4b=6c ((3a)/(12))=((4b)/(12))=((6c)/(12)) (a/4)=(b/3)=(c/2)=k (let) ∴a=4k, b=3k,c=2k a+b+c=27 ⇒4k+3k+2k=27 ⇒9k=27⇒k=3 a−c=4k−2k=2k=2×3=6

Answered by Rasheed.Sindhi last updated on 25/Oct/22

(1/(3a)) = (1/(4b)) = (1/(6c)) ∧ a+b+c=27; a−c=? (1/(3a)) = (1/(4b)) = (1/(6c))⇒((4×1)/(4(3a))) = ((3×1)/(3(4b))) = ((2×1)/(2(6c))) =(9/(12(a+b+c)))=(3/(4×27))=(1/(36)) 3a=36 ∧ 6c=36 3a=36 ∧ 3c=18 ⇒3a−3c=18 ⇒a−c=6