Question 542689: twelve years ago, a man was three times as old as his daughter. nine years from now, the man will be twice as old as his son. four years from now, the sum of the ages of the son and daughter will equal the age of the man. how old is each now?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Let m = man's present age
Let d = daughter's
Let s = son's
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Write an equation for each statement, simplify:
:
twelve years ago, a man was three times as old as his daughter.
m - 12 = 3(d - 12)
m - 12 = 3d - 36
m = 3d - 36 + 12
m = 3d - 24
:
" nine years from now, the man will be twice as old as his son."
m + 9 = 2(s + 9)
m + 9 = 2s + 18
m = 2s + 18 - 9
m = 2s + 9
:
"four years from now, the sum of the ages of the son and daughter will equal the age of the man."
m + 4 = s+4 + d+4
m + 4 = s + d + 8
m = s + d + 8 - 4
m = s + d + 4
replace m with (3d-24)
3d - 24 = s + d + 4
3d - d - s = 4 + 24
2d - s = 28
then in the same equation
m = s + d + 4
replace m with (2s + 9)
2s + 9 = s + d + 4
2s = s + d + 4 - 9
-d + 2s - s = - 5
-d + s = -5
:
Use elimination on these two equations
2d - s = 28
-d + s = -5
---------------addition eliminates s, find d
d = 23 is the daughters age
:
Find s using the equation -d + s = -5
-23 + s = -5
s = -5 + 23
s = 18 is the son's age
:
Use m = 3d - 24 to find the man's
m = 3(23) - 24
m = 69 - 24
m = 45 is the man's age
:
Check this in the equation: m = s + d + 4
45 = 18 + 23 + 4
45 = 45
:
:
how old is each now? man is 45, daughter is 23, son is 18
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