Question 171937: ten years ago, a man was 3 times as old as his son. in 6 years, he will be twice as old as his son. how old is each now? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=son's age now;
x-10=son's age ten years ago;
x+6=son's age in 6 years
And let y=man's age now;
y-10=man's age ten years ago;
y+6=man's age in 6 years
Now we are told the following:
y-10=3(x-10)=
y-10=3x-30=
3x-y=20----------------eq1
and
y+6=2(x+6)=
y+6=2x+12=
2x-y=-6------------------eq2
subtract eq2 from eq1:
x=26----------------------------son's age now
substitute x=26 into eq1:
3*26-y=20=
78-y=20
-y=-58
y=58-------------------man's age now
CK
58-10=3(26-10)
48=48
and
58+6=2(26+6)
64=64
Hope this helps---ptaylor
