Question 582513
Let {{{ a }}} = number of $12 ones she sold
Let {{{ b }}} = number of $13 ones she sold
Let {{{ c }}} = number of $22 ones she sold
given:
(1) {{{ a = b - 1 }}}
(2) {{{ a + b + c = 40 }}}
(3) {{{ 12a + 13b + 22c = 643 }}}
This is 3 equations and 3 unknowns, so it's solvable
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Multiply both sides of (2) by {{{22}}} and
subtract (3) from (2)
(2) {{{ 22a + 22b + 22c = 880 }}}
(3) {{{ -12a - 13b - 22c = -643 }}}
{{{ 10a + 9b = 237 }}}
Substitute (1) into this result
{{{ 10*( b - 1 ) + 9b = 237 }}}
{{{ 10b - 10 + 9b = 237 }}}
{{{ 19b = 247 }}}
{{{ b = 13 }}}
and, since
(1) {{{ a = b - 1 }}}
(1) {{{ a = 13 - 1 }}}
(1) {{{ a = 12 }}}
and
(2) {{{ a + b + c = 40 }}}
(2) {{{ 12 + 13 + c = 40 }}}
(2) {{{ 25 + c = 40 }}}
(2) {{{ c = 40 - 25 }}}
(2) {{{ c = 15 }}}
She sold 15 of the $22 subscriptions
check answer:
(3) {{{ 12a + 13b + 22c = 643 }}}
(3) {{{ 12*12 + 13*13 + 22*15 = 643 }}}
(3) {{{ 144 + 169  + 330 = 643 }}}
(3) {{{ 643 = 643 }}}
OK