Question 825211
Let {{{ j }}} = Jacob's age now
Let {{{ b }}} = brother's age now
{{{ j - 5 }}} = Jacob's age 5 years ago
{{{ b - 5 }}} = brother's age 5 years ago
{{{ j + 3 }}} = Jacob's age in 3 years
{{{ b + 3 }}} = brother's age in 3 years
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(1) {{{ j - 5 = (1/6)*( b - 5 ) }}}
(2) {{{ 2*( j+3 ) = b + 3 }}}
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(1) {{{ 6*( j-5 ) = b - 5 }}}
(1) {{{ 6j - 30 = b - 5 }}}
(1) {{{ 6j - b = 25 }}}
and
(2) {{{ 2j + 6 = b + 3 }}}
(2) {{{ 2j - b = -3 }}}
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Subtract (2) from (1)
(1) {{{ 6j - b = 25 }}}
(2) {{{ -2j + b = 3 }}}
{{{ 4j = 28 }}}
{{{ j = 7 }}}
Jacob is 7 now
check:
(1) {{{ j - 5 = (1/6)*( b - 5 ) }}}
(1) {{{ 7 - 5 = (1/6)*( b - 5 ) }}}
(1) {{{ 2 = (1/6)*( b-5 ) }}}
(1) {{{ 12 = b - 5 }}}
(1) {{{ b = 17 }}}
and
(2) {{{ 2*( j+3 ) = b + 3 }}}
(2) {{{ 2*( 7+3 ) = 17 + 3 }}}
(2) {{{ 2*10 = 20 }}}
(2) {{{ 20 = 20 }}}
OK