SOLUTION: Two years ago Philip was three times as old as his son.
Two years hence, twice his age will be equal to five times
that of his son. Find their present ages.
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-> SOLUTION: Two years ago Philip was three times as old as his son.
Two years hence, twice his age will be equal to five times
that of his son. Find their present ages.
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Question 1031190: Two years ago Philip was three times as old as his son.
Two years hence, twice his age will be equal to five times
that of his son. Find their present ages. Found 2 solutions by josmiceli, Edwin McCravy:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = father's age now
Let = his son's age now
-----------------------------
(1)
(2)
------------------------
(1)
(1)
and
(2)
(2)
----------------------
Multiply both sides of (1) by
and add the equations
(1)
(2)
--------------------
and
(1)
(1)
(1)
(1)
--------------------
The son is 14 and the father is 38
-----------------------------
check:
(1)
(1)
(1)
(1)
OK
(2)
(2)
(2)
(2)
OK
Two years ago Philip was three times as old as his son.
Two years hence, twice his age will be equal to five times
that of his son. Find their present ages.
Let Philip's present age be P
Let his son's present age be S
Two years ago Philip was...
Two years ago Philip was P-2
Two years ago his son was S-2
Two years ago Philip was three times as old as his son.
P-2 = 3(S-2)
P-2 = 3S-6
P-3S = -4
Two years hence,
Two years hence Philip will be P+2
Two years hence his son will be will be S+2
Two years hence, twice his age will be equal to five
times that of his son.
2(P+2) = 5(S+2)
2P+4 = 5S+10
2P-5S = 6
Solve the system either by substitution or elimination:
You finish.
Edwin
