Question 526398
Let {{{a}}} = number of buyers 35 and younger
Let {{{b}}} = number of buyers 36-59
Let {{{c}}} = number of buyers 60 and older
given:
(1) {{{ a + b + c = 100000 }}}
(2) {{{ b / ( a + b + c )  + c / ( a + b + c ) = a / ( a + b + c ) + .32 }}}
(3) {{{ (2c) / ( a + b + c ) = b / ( a + b + c ) - .21 }}}
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There are 3 equations and 3 unknowns, so it's solvable
Substitute (1) into (2)
(2) {{{ b / 100000  + c / 100000 = a / 100000 + .32 }}}
(2) {{{ b + c = a + .32*100000 }}}
(2) {{{ b + c = a + 32000 }}}
Substitute (1) into (3)
(3) {{{ (2c) / 100000 = b / 100000 - .21 }}}
(3) {{{ 2c = b - 21000 }}}
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(3) {{{ b = 2c + 21000 }}}
Substitute (3) into (2)
(2) {{{ 2c + 21000 + c = a + 32000 }}}
(2) {{{ a = 3c + 21000 - 32000 }}}
(2) {{{ a = 3c - 11000 }}}
Substitute (2) and (3) into (1)
(1) {{{ a + b + c = 100000 }}}
1) {{{ 3c - 11000 + 2c + 21000 + c = 100000 }}}
(1) {{{ 6c = 100000 + 11000 - 21000 }}}
(1) {{{ 6c = 90000 }}}
(1) {{{ c = 15000 }}}
and, since
(2) {{{ a = 3c - 11000 }}}
(2) {{{ a = 3*15000 - 11000 }}}
(2) {{{ a = 45000 - 11000 }}}
(2) {{{ a = 34000 }}}
and
(3) {{{ b = 2c + 21000 }}}
(3) {{{ b = 2*15000 + 21000 }}}
(3) {{{ b = 30000 + 21000 }}}
(3) {{{ b = 51000 }}}
check:
(1) {{{ a + b + c = 100000 }}}
(1) {{{ 34000 + 51000 + 15000 = 100000 }}}
(1) {{{ 100000 = 100000 }}}
{{{ 34000/100000 = .34 }}} = 34% of buyers are 35 and younger
{{{ 51000/100000 = .51 }}} = 51% of buyers are 36-59
{{{ 15000/100000 = .15 }}} = 15% of buyers are 60 and older