Question 741287
Let {{{ a }}} = teacher's age
Let {{{ b }}} = student's age
{{{ a + 16 }}} = teacher's age in {{{ 16 }}} years
{{{ b + 16 }}} = student's age in {{{ 16 }}} years
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(1) {{{ a = 4b }}}
(2) {{{ b + 16 = (1/2)*( a + 16 ) }}}
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This is 2 equations and 2 unknowns, so it's solvable
(1) {{{ b = a/4 }}}
Substitute (1) into (2)
(2) {{{ a/4 + 16 = (1/2)*( a + 16 ) }}}
Multiply both sides by {{{ 4 }}}
(2) {{{ a + 64 = 2*( a + 16 ) }}}
(2) {{{ a + 64 = 2a + 32 }}}
(2) {{{ a = 64 - 32 }}}
(2) {{{ a = 32 }}}
and, since
(1) {{{ b = a/4 }}}
(1) {{{ b = 32/4 }}}
(1) {{{ b = 8 }}}
The teacher is 32 and the student is 8
check:
(2) {{{ b + 16 = (1/2)*( a + 16 ) }}}
(2) {{{ 8 + 16 = (1/2)*( 32 + 16 ) }}}
(2) {{{ 24 = (1/2)*48 }}}
(2) {{{ 48 = 48 }}}
OK