Question 242951: in four years time, a father will be three times as old as his son. six years ago he was seven times as old as his son. how old are they now ?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! x = father's current age
y = son's current age
In 4 years, x will be 3 times y, which is shown as: x+4 = 3(y+4)
Six years ago (x-6) he was 7 times as old: x-6 = 7(y-6)
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Summarizing what we know:
x + 4 = 3(y+4) = 3y + 12
x - 6 = 7(y-6) = 7y - 42
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Deriving a value for x:
x + 4 = 3y + 12
x = 3y + 12 - 4
x = 3y + 8
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Substituting into the second equation,
x - 6 = 7y - 42
3y+8 -6 = 7y -42
3y+2 = 7y - 42
2+42 = 7y-3y
44 = 4y
4y = 44
y = 11
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Going back to a formula we know
x = 3y + 8
x = 3(11) + 8 = 41
x = 41
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Check our work by checking the other facts given.
How old will they be in 4 years?
x+4 = 41+4 = 45
y+4 = 11+4 = 15
Does 15 = 3*15? Yes, it does.
How old were they 6 years ago?
x-6 = 41-6 = 35
y-6 = 11-6 = 5
Is x = 7 times the son's age?
35 = 7*5? Yes.
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So, we are confident the current ages are:
father = 41
son = 11
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