Question 1012706
Let {{{ a }}} =number of 1-point shots
Let {{{ b }}} =number of 2-point shots
Let {{{ c }}} =number of 3-point shots
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(1) {{{ a + b + c = 355 }}}
(2) {{{ 1*a + 2*b + 3*c = 646 }}}
(3) {{{ b = c + 27 }}}
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This is 3 equations and 3 unknowns, 
so it is solvable
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Substitute (3) into (2)
(2) {{{ 1*a + 2*( c + 27 ) + 3*c = 646 }}}
(2) {{{ a + 2c + 54 + 3c = 646 }}}
(2) {{{ a + 5c = 592 }}}
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Substitute (3) into (1)
(1) {{{ a +c + 27 + c = 355 }}}
(1) {{{ a + 2c = 328 }}}
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Subtract (1) from (2)
(2) {{{ a + 5c = 592 }}}
(1) {{{ -a - 2c = -328 }}}
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{{{ 3c = 264 }}}
{{{ c = 88 }}}
and
(3) {{{ b = c + 27 }}}
(3) {{{ b = 88 + 27 }}}
(3) {{{ b = 115 }}}
and 
(1) {{{ a + b + c = 355 }}}
(1) {{{ a + 115 + 88 = 355 }}}
(1) {{{ a = 355 - 203 }}}
(1) {{{ a = 152 }}}
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88 3-point goals
115 2-point goals
152 1-point free throws
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check:
(2) {{{ 1*a + 2*b + 3*c = 646 }}}
(2) {{{ 1*152 + 2*115 + 3*88 = 646 }}}
(2) {{{ 152 + 230 + 264 = 646 }}}
(2) {{{ 646 = 646 }}}
You can check other 2 equations