Question 1039437
Let {{{ a }}} = number of packages of gum Terry bought
Let {{{ b }}} = number of mints
Let {{{ c }}} = number of candy bars
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(1) {{{ a = b + 1 }}}
(2) {{{ b = 3c }}}
(3) {{{ 24a + 10b + 35c = 572 }}} ( in cents )
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There are 3 equations and 3 unknowns, so it's solvable
(2) {{{ c = b/3 }}}
Substitute (2) into (3)
(3) {{{ 24a + 10b + 35*( b/3) = 572 }}}
Substitute (1) into (3)
(3) {{{ 24*( b + 1 ) + 10b + 35*( b/3 ) = 572 }}}
(3) {{{ 24b + 24 + 10b + (35/3)*b = 572 }}}
Multiply both sides by {{{ 3 }}}
(3) {{{ 3*34b + 3*24 + 35b = 3*572 }}}
(3) {{{ 102b + 72 + 35b = 1716 }}}
(3) {{{ 137b = 1644 }}}
(3) {{{ b = 12 }}}
and
(1) {{{ a = b + 1 }}}
(1) {{{ a = 12 + 1 }}}
(1) {{{ a = 13 }}}
and
(2) {{{ c = b/3 }}}
(2) {{{ c = 12/3 }}}
(2) {{{ c = 4 }}}
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He bought 13 pkgs of gum
He bought 12 mints
He bought 4 candy bars
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check:
(3) {{{ 24a + 10b + 35c = 572 }}} ( in cents )
(3) {{{ 24*13 + 10*12 + 35*4 = 572 }}} 
(3) {{{ 312 + 120 + 140 = 572 }}}
(3) {{{ 572 = 572 }}}
OK