Question 186598
Let {{{f}}} = Father's age now
Let {{{s}}} = Son's age now
given:
{{{f + s = 55}}}
The Son's age in 2 years is
{{{s + 2}}}
When the Father was twice as old as the Son will be in two years is
{{{2*(s + 2)}}}
How old was the Son then?
The difference in their ages is a constant
{{{f - s = c}}}
{{{s = f - c}}}
{{{2*(s + 2) - c}}}
{{{2s + 4 - c}}}
The Father's present age is three time the Son's at that time
{{{f = 3*2*(s + 2) - 3c}}}
(1) {{{f = 6s + 12 - 3c}}}
also,
(2) {{{f + s = 55}}}
(3) {{{f - s = c}}}
Substitute (3) in (1)
(1) {{{f = 6s + 12 - 3*(f - s)}}}
(1) {{{f = 6s + 12 - 3f + 3s}}}
(1) {{{4f - 9s = 12}}}
Multiply both sides of (2) by {{{4}}} and 
subtract (1) from (2)
(2) {{{4f + 4s = 220}}}
(1) {{{-4f + 9s = -12}}}
(3) {{{13s = 208}}}
(3) {{{s = 16}}}
and, since
(2) {{{f + s = 55}}}
(2) {{{f = 55 - 16}}}
(2) {{{f = 39}}}
The father is 39 now, and the Son is 16
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check answer:
The Son's age in 2 years is
{{{16 + 2 = 18}}}
When the Father was twice as old as the Son will be in two years is
{{{2*(16 + 2)}}}
{{{36}}}
How old was the Son then?
The difference in their ages is a constant
{{{36 - s = c}}}
{{{s = 36 - c}}}
{{{2*(s + 2) - c}}}
{{{2s + 4 - c}}}
The Father's present age is three time the Son's at that time
{{{39 = 3*2*(s + 2) - 3c}}}
(1) {{{39 = 6*16 + 12 - 3c}}}
(1) {{{39 = 96 + 12 - 3c}}}
(1) {{{3c = 96 + 12 - 39}}}
(1) {{{3c = 108 - 39}}}
(1) {{{3c = 69}}}
(1) {{{c = 23}}}
and
(3) {{{f - s = c}}}
(3) {{{39 - 16 = 23}}}
OK