Find the 'age' using algebra

[MathsFormula]

A man is 32 years older than his son.
Ten years ago he was three times as old as his son was then. Find the present age of each.

Let the son's age be X

So the man's age now is (32+X)

So 10 years ago his age was (32+X)-10

He was then three times as old as his son
was then so :

(32+X)-10 = 3X

22 = 2X

X = 11 = age of son

Age of father is 32 + 11 = 43

WRONG according to book that says answer is 26 and 58

Please advise where I've gone wrong. Thank you

 

[Deleted member 4993]

A man is 32 years older than his son.
Ten years ago he was three times as old as his son was then. Find the present age of each.

Let the son's age be X

So the man's age now is (32+X)

So 10 years ago his age was (32+X)-10

He was then three times as old as his son
was then so :

(32+X)-10 = 3(X - 10) ............... Now continue

Ten years ago, Son.s age was (x - 10)

 

[stapel]

First, let me say "Thank you!" for having shown your work and reasoning so nicely!

A man is 32 years older than his son.
Ten years ago he was three times as old as his son was then. Find the present age of each.

Let the son's age be X

So the man's age now is (32+X)

So 10 years ago his age was (32+X)-10

He was then three times as old as his son
was then so :

(32+X)-10 = 3X

Almost. Since X means "the son's age now", then 3X means "three times the son's current age". But you want "three times the son's age back ten years ago". So translate each piece explicitly:

. . .son's age now: x
. . .father's age now: x + 32

. . .father's age, ten years ago: (x + 32) - 10 = x + 22
. . .son's age, ten years ago: x - 10
. . .three times son's age, ten years ago: 3(x - 10)

Then translate "(the father's age, then years ago) was (three times) (the son's age, ten years ago)"

. . .x + 22 = 3(x - 10)

See what that gets you.

 

[MathsFormula]

Thanks for the ideas. Will go through it again.

Just been through it. Got the answer. Thanks.