SOLUTION: six years ago a man was four times as old as his son was then. in twelve years, the many will be just twice as old as his son will be then. Find their present ages.

Algebra ->  Equations -> SOLUTION: six years ago a man was four times as old as his son was then. in twelve years, the many will be just twice as old as his son will be then. Find their present ages.      Log On


   

Question 1130667: six years ago a man was four times as old as his son was then. in twelve years, the many will be just twice as old as his son will be then. Find their present ages.
Found 2 solutions by Boreal, josmiceli:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
son is x now
x-6 was his age 6 years ago
Man's age 6 years ago was 4(x-6) or 4x-24. That means the man is 4x-18 years old now.
In 12 years, the man will be 4x-6 years old and that is twice as old as his son in 12 years, which is x+12
so 4x-6=2(x+12)
4x-6=2x+24
2x=30
x=15
son is 15 now and when he was 9, his man was 36 so
man is 42 now
15 and 42 ANSWER
check
in 12 years, the man will be 54 and the son 27, which is twice as old.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = man's age now
Let +b+ = son's age now
---------------------------------
(1) +a+-+6+=+4%2A%28+b+-+6+%29+
(2) +a+%2B+12+=+2%2A%28+b+%2B+12+%29+
---------------------------------
(1) +a+-+6+=+4b+-+24+
(1) +-a+%2B+4b+=+18+
and
(2) +a+%2B+12+=+2b+%2B+24+
(2) +a+-+2b+=+12+
Add the equations
+2b+=+30+
+b+=+15+
and
(1) +a+-+6+=+4%2A%28+15+-+6+%29+
(1) +a+-+6+=+36+
(1) +a+=+42+
---------------------
The man is 42
His son is 15
---------------------
check:
(1) +a+-+6+=+4%2A%28+b+-+6+%29+
(1) +42+-+6+=+4%2A%28+15+-+6+%29+
(1) +36+=+4%2A9+
(1) +36+=+36+
OK
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