Question 603784: sonny is twice as old as david. Four years ago, he was four times as old as david. When will the sum of their ages be 66? Found 3 solutions by mananth, josmiceli, ikdeep:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! David be x years
Sonny = 2x
4 years ago their ages
David = x-4
Sonny = 2x-4
2x-4=4(x-4)
2x-4 = 4x-16
4x-2x=16-4
2x=12
/2
x= 6 --- David's age
Sonny's age = 12
Let their sum of ages be 66 after y years
6+y+12+y = 66
2y+18=66
2y=66-18
2y= 48
/2
y=24
their sum will be 66 after 24 years
CHECK
David =6 + 24= 30
Sonny = 12+24= 36
Total = 66
You can put this solution on YOUR website! Let = Sonny's age now
Let = David's age now
given:
(1)
(2)
-------------------------------
Let = the number of years to add to
each one's age, so the sum =
(3)
------------------------------
This is 3 equations and 3 unknowns, so it's solvable
(1)
(1)
Substitute (1) into (2)
(2)
(2)
(2)
(2)
and, since
(1)
(1)
(1)
Substitute these results into (3)
(3)
(3)
(3)
(3)
(3)
In 24 years the sum of their ages will be 66
check:
(3)
(3)
(3)
(3)
OK
You can put this solution on YOUR website! In 24 years the sum of both the ages = 66
You can check step by step solution from the below link
http://www.algebraden.com/solved-problems/find-years/32.htm
Hope this will help you.
