Question 1042468
You need a total of 45 dinners
Let {{{ a }}} = number of chicken dinners ordered
Let {{{ b }}} = number of salmon dinners ordered
Let {{{ c }}} = number of steak dinners ordered
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(1) {{{ a = (1/2)*( b + c ) }}}
(2) {{{ a + b + c = 45 }}}
(3) {{{ 9a + 12b + 15c = 525 }}}
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(1) {{{ 2a = b + c }}}
(1) {{{ c = 2a - b }}}
Substitute (1) into (2)
(2) {{{ a + b + 2a - b = 45 }}}
(2) {{{ 3a  = 45 }}}
(2) {{{ a = 15 }}}
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(1) {{{ c = 2a - b }}}
(1) {{{ c = 2*15 - b }}}
(1) {{{ c = 30 - b }}}
Substitute (1) into (3)
(3) {{{ 9a + 12b + 15c = 525 }}}
(3) {{{ 9*15 + 12b + 15*( 30 - b ) = 525 }}}
(3) {{{ 135 + 12b + 450 - 15b = 525 }}}
(3) {{{ -3b = 525 - 585 }}}
(3) {{{ 3b = 60 }}}
(3) {{{ b = 20 }}}
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(1) {{{ c = 2a - b }}}
(1) {{{ c = 2*15 - 20 }}}
(1) {{{ c = 30 - 20 }}}
(1) {{{ c = 10 }}}
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15 chicken dinners were ordered
20 salmon dinners were ordered
10 steak dinners were ordered
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check:
(1) {{{ a = (1/2)*( b + c ) }}}
(1) {{{ 15 = (1/2)*( 20 + 10 ) }}}
(1) {{{ 15 = 30/2 }}}
(1) {{{ 15 = 15 }}}
and
(2) {{{ a + b + c = 45 }}}
(2) {{{ 15 + 20 + 10 = 45 }}}
(2) {{{ 45 = 45 }}}
and
(3) {{{ 9a + 12b + 15c = 525 }}}
(3) {{{ 9*15 + 12*20 + 15*10 = 525 }}}
(3) {{{ 135 + 240 + 150 = 525 }}}
(3) {{{ 525 = 525 }}}
OK