five years ago beth s age was three times that of amy s age ten years ago beth s age

(B-5) = 3(A-5) or B = 3A-10
(B-10) = 1.5(C-10)or B = 1.5C-5

so from the above 2 equations,
3A-10 = 1.5C-5
or A = (3C+10)/6
10 years agoHelpfull: Yes(2) No(3)
Beth's prag age=x.
Amy's prag age=y.
AQ,
X-5=3(y-5).......1
And,
X-10=(1/2)*(c-10)......2
Solving 1&2 ,
Y=3/2c+5....
9 years agoHelpfull: Yes(2) No(1)
5 years ago beth's age=3x.
Amy's age=x.
AQ,
10 yers ago beth's was 1/2 that of chelsea
so, 3x-5=1/2(c-10)
x=c/6(this is Amy's age 5 years ago)
amy's present age x=(c/6)+5
x=(c+30)/6
9 years agoHelpfull: Yes(2) No(1)
given, (beth - 5)=3*(amy-5)
=>beth=3*amy - 10
also given, (beth-10)=(chelsea-10)/2
=>so chelsea(c)=6a-30
=>amy's age = (c/6+5)or (c+30)/5
10 years agoHelpfull: Yes(1) No(2)
(B-5) = 3(A-5)
(B-10) = 0.5(C-10)or B = 0.5C-5+10=0.5c+5

so from the above 2 equations,
b-5=3(a-5)
substitute b in above line
0.5c+5=3(a-5)
0.5c+20=3a
a=0.5c+20/3
or A = (3C+10)/6
10 years agoHelpfull: Yes(1) No(0)
Present age of Beth,Amy and Chelsea is y,x and c resp.
so,(y-5)=3(x-5)..........................(i)
and(y-10)=1/2(c-10)
on solving we get y=(c+10)/2
from..........(i)
((c+10)/2)-5)=(3x-15)
on solving x=((c+30)/6) years.

THANK YOU.
9 years agoHelpfull: Yes(1) No(0)

five years ago,beth's age was three times that of amy's age.ten years ago,beth's age was one half that of chelsea.if c represents chelsea's current age,which of the following represents amy's current age