Question 841466
Let {{{ a }}} = number of $20 checks
Let {{{ b }}} = number of $50 checks
Let {{{ c }}} = number of $100 checks
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(1) {{{ a + b + c = 10 }}}
(2) {{{ 20a + 50b + 100c = 370 }}}
(3) {{{ a = 2b }}}
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there are 3 equations and 3 unknowns,
so it's solvable
Multiply both sides of (1) by {{{ 100 }}}
and subtract (2) from (1)
(1) {{{ 100a + 100b + 100c = 1000 }}}
(2) {{{ -20a - 50b - 100c = -370 }}}
{{{ 80a + 50b = 630 }}}
Substitute (3) into this result
{{{ 80*2b + 50b = 630 }}}
{{{ 210b = 630 }}}
{{{ b = 3 }}}
and, since
(3) {{{ a = 2b }}}
(3) {{{ a = 6 }}}
and
(1) {{{ a + b + c = 10 }}}
(1) {{{ 6 + 3 + c = 10 }}}
(1) {{{ c = 1 }}}
There are 6 $20 checks
There are 3 $50 checks
There is 1 $100 check
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check:
(2) {{{ 20a + 50b + 100c = 370 }}}
(2) {{{ 2a + 5b + 10c = 37 }}}
(2) {{{ 2*6 + 5*3 + 10*1 = 37 }}}
(2) {{{ 12 + 15 + 10 = 37 }}}
(2) {{{ 37 = 37 }}}
OK