Question 603784
Let {{{ s }}} = Sonny's age now
Let {{{ d }}} = David's age now
given:
(1) {{{ s = 2d }}}
(2) {{{ s - 4 = 4*( d - 4 ) }}}
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Let {{{ n }}} = the number of years to add to 
each one's age, so the sum = {{{ 66 }}}
(3) {{{ s + n + d + n = 66 }}}
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This is 3 equations and 3 unknowns, so it's solvable
(1) {{{ s = 2d }}}
(1) {{{ d = s/2 }}}
Substitute (1) into (2)
(2) {{{ s - 4 = 4*( s/2 - 4 ) }}}
(2) {{{ s - 4 = 2s - 16 }}}
(2) {{{ s = 16 - 4 }}}
(2) {{{ s = 12 }}}
and, since
(1) {{{ d = s/2 }}}
(1) {{{ d = 12/2 }}}
(1) {{{ d = 6 }}}
Substitute these results into (3)
(3) {{{ 12 + n + 6 + n = 66 }}}
(3) {{{ 2n + 18 = 66 }}}
(3) {{{ 2n = 66 - 18 }}}
(3) {{{ 2n = 48 }}}
(3) {{{ n = 24 }}}
In 24 years the sum of their ages will be 66
check:
(3) {{{ s + n + d + n = 66 }}}
(3) {{{ 12 + 24 + 6 + 24 = 66 }}}
(3) {{{ 36 + 30 = 66 }}}
(3) {{{ 66 = 66 }}}
OK